Geometry

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Name the three undefined terms of geometry.
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   Point, line and plane

2. What does \( \overline{PQ} \) mean?
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   The segment with endpoints \( P \) and \( Q \)

3. What does \( PQ \) mean, without the bar?
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   The length of that segment, a number

4. Are \( \overrightarrow{XY} \) and \( \overrightarrow{YX} \) the same ray?
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   They have different endpoints and point in opposite directions. No

5. How many lines pass through two distinct points?
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   Exactly one

6. How many segments are determined by five points, no three collinear?
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   A segment is determined by a pair of points, so count pairs. The first point can be chosen five ways and the second four, giving 20 ordered pairs, but a segment has no direction so each is counted twice: \( 20 \div 2 = 10 \). Checking by listing with points \( A \) through \( E \): \( AB, AC, AD, AE \) is four; \( BC, BD, BE \) is three; \( CD, CE \) is two; \( DE \) is one. Total \( 4 + 3 + 2 + 1 = 10 \). 10

7. Three points lie on a line. How many planes contain all three?
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   The postulate that three points determine exactly one plane requires them to be noncollinear, so it does not apply here. Picture the line as an axle: a plane containing it can be rotated about it to any position and still contains all three points. Infinitely many

8. Explain why point, line and plane are left undefined rather than being defined carefully.
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   Every definition explains one term using other terms. Those other terms then need definitions of their own, using still others. Only two endings are possible: the chain runs forever, or it comes back on itself and defines a term using something that was defined from it. Neither gives a foundation a proof can rest on. The resolution is to stop deliberately. A few terms are taken as understood without definition, described well enough to work with, and every other term in the subject is defined from them. That gives a finite chain with a bottom. The same structure appears in the statements: a few postulates are accepted without proof, and every theorem is proved from them. Undefined terms and postulates are the same idea applied to words and to claims. Defining every term would require an infinite or circular chain, so a few are taken as understood

9. Point \( B \) is between \( A \) and \( C \). Which of \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), \( \overrightarrow{BA} \), \( \overrightarrow{BC} \) name the same ray?
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   A ray is determined by its endpoint and its direction, not by which second point is named. \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) both start at \( A \) and head toward \( B \) and \( C \), which lie the same way since \( B \) is between them. Same endpoint, same direction, so they are the same ray. \( \overrightarrow{BA} \) starts at \( B \) heading toward \( A \). \( \overrightarrow{BC} \) starts at \( B \) heading the opposite way toward \( C \). Same endpoint, opposite directions, so they are different rays. Together they make the whole line, and two such rays are called opposite rays. \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are the same; \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \) are opposite rays

10. A student writes \( \overline{AB} = \overline{CD} \) in a proof. Explain what is wrong and give both correct versions.
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    The equals sign relates numbers, and \( \overline{AB} \) is not a number; it is a segment, a set of points. Saying one set of points equals another would mean they are the same segment, which is almost never what is intended. Two correct statements are available, and they say different things in different language. \( \overline{AB} \cong \overline{CD} \) says the segments are congruent. The reason column would cite a theorem about figures. \( AB = CD \) says their lengths are equal. The reason column would cite an arithmetic property, and this is the form needed before adding or subtracting lengths. They are equivalent, and the definition of congruent segments is the theorem that licenses moving between them. A proof that needs to add lengths converts to the equation form first and says so. Use \( \overline{AB} \cong \overline{CD} \) for the figures or \( AB = CD \) for the lengths; an equals sign between two segments is not meaningful

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. If \( B \) is between \( A \) and \( C \), \( AB = 7 \) and \( BC = 11 \), find \( AC \).
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   Segment addition postulate. 18

2. Find the midpoint of the segment from \( (2, 6) \) to \( (10, 4) \).
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   Average each coordinate. \( (6, 5) \)

3. Find the distance between \( (0, 0) \) and \( (3, 4) \).
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   \( \sqrt{9 + 16} = 5 \). 5

4. \( M \) is the midpoint of \( \overline{AB} \) and \( AM = 9 \). Find \( AB \).
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   A midpoint makes two congruent segments. 18

5. What postulate lets you write \( AB + BC = AC \)?
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   The segment addition postulate, provided \( B \) is between \( A \) and \( C \)

6. \( B \) is between \( A \) and \( C \) with \( AB = 4x - 3 \), \( BC = x + 5 \) and \( AC = 32 \). Find all three lengths.
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   By the segment addition postulate, \( (4x - 3) + (x + 5) = 32 \), so \( 5x + 2 = 32 \), giving \( 5x = 30 \) and \( x = 6 \). \( AB = 4(6) - 3 = 21 \) and \( BC = 6 + 5 = 11 \). Check: \( 21 + 11 = 32 \). Correct, and both parts are positive. \( AB = 21 \), \( BC = 11 \), \( AC = 32 \)

7. Find the distance between \( (-2, 5) \) and \( (4, -3) \).
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   Horizontal difference: \( 4 - (-2) = 6 \). Vertical difference: \( -3 - 5 = -8 \). \( d = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \). The negative vertical difference is squared away, which is why the order of subtraction does not matter. 10

8. \( M(3, -1) \) is the midpoint of \( \overline{AB} \) and \( A \) is \( (7, 5) \). Find \( B \).
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   The midpoint formula gives two equations. For the \( x \) coordinates: \( \dfrac{7 + x}{2} = 3 \), so \( 7 + x = 6 \) and \( x = -1 \). For the \( y \) coordinates: \( \dfrac{5 + y}{2} = -1 \), so \( 5 + y = -2 \) and \( y = -7 \). So \( B \) is \( (-1, -7) \). Check by averaging: \( \dfrac{7 + (-1)}{2} = 3 \) and \( \dfrac{5 + (-7)}{2} = -1 \), which is \( M \). Correct. A useful shortcut worth knowing: the midpoint is as far from \( B \) as from \( A \), so from \( A(7,5) \) move to \( M(3,-1) \) by going 4 left and 6 down, then repeat the same move to reach \( (-1, -7) \). \( B(-1, -7) \)

9. Explain why the distance formula is the Pythagorean theorem.
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   Take two points \( (x_1, y_1) \) and \( (x_2, y_2) \). Draw the horizontal segment from the first to \( (x_2, y_1) \) and the vertical segment from there to the second. Those two segments meet at a right angle, because one is horizontal and the other vertical, so together with the segment joining the original points they form a right triangle. The horizontal leg has length \( |x_2 - x_1| \) and the vertical leg \( |y_2 - y_1| \). The segment joining the two points is the hypotenuse. The Pythagorean theorem gives \( d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \), and taking the positive square root gives the distance formula. The absolute values disappear because each quantity is squared. This is worth knowing rather than memorizing the formula, because a forgotten formula can be rebuilt in ten seconds from a sketch. The horizontal and vertical differences are the legs of a right triangle whose hypotenuse is the distance

10. \( B \) is between \( A \) and \( C \). \( AB = 2x + 1 \), \( BC = 3x - 4 \), and \( AB = BC \). Find \( AC \), and say what this makes \( B \).
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    Two facts are given, so use them in order. First, \( AB = BC \) gives \( 2x + 1 = 3x - 4 \), so \( x = 5 \). Then \( AB = 2(5) + 1 = 11 \) and \( BC = 3(5) - 4 = 11 \). Equal, as required. By the segment addition postulate, \( AC = 11 + 11 = 22 \). Since \( B \) lies between \( A \) and \( C \) and divides \( \overline{AC} \) into two congruent segments, \( B \) is the midpoint of \( \overline{AC} \) by the definition of a midpoint. Both conditions in the definition matter and both were checked: a point equidistant from \( A \) and \( C \) but off the line is not the midpoint, and betweenness is what rules that out. \( AC = 22 \), and \( B \) is the midpoint of \( \overline{AC} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Classify an angle measuring \( 143^\circ \).
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   Obtuse

2. What is the vertex of \( \angle PQR \)?
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   The middle letter. \( Q \)

3. \( \overrightarrow{BD} \) bisects \( \angle ABC \) and \( m\angle ABC = 84^\circ \). Find \( m\angle ABD \).
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   A bisector makes two congruent angles. \( 42^\circ \)

4. \( m\angle ABD = 25^\circ \) and \( m\angle DBC = 40^\circ \), with \( D \) interior. Find \( m\angle ABC \).
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   Angle addition postulate. \( 65^\circ \)

5. What is the difference between \( \angle ABC \) and \( m\angle ABC \)?
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   The first is the figure; the second is its measure, a number

6. \( \overrightarrow{BD} \) is in the interior of \( \angle ABC \) with \( m\angle ABD = 3x + 5 \), \( m\angle DBC = x + 15 \) and \( m\angle ABC = 88^\circ \). Find all three.
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   By the angle addition postulate, \( (3x + 5) + (x + 15) = 88 \), so \( 4x + 20 = 88 \), giving \( 4x = 68 \) and \( x = 17 \). \( m\angle ABD = 3(17) + 5 = 56^\circ \) and \( m\angle DBC = 17 + 15 = 32^\circ \). Check: \( 56 + 32 = 88 \). Correct. \( 56^\circ \), \( 32^\circ \) and \( 88^\circ \)

7. \( \overrightarrow{BD} \) bisects \( \angle ABC \), with \( m\angle ABD = 5x - 4 \) and \( m\angle DBC = 3x + 10 \). Find \( m\angle ABC \).
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   A bisector makes the two parts congruent, so their measures are equal: \( 5x - 4 = 3x + 10 \), giving \( 2x = 14 \) and \( x = 7 \). Each part measures \( 5(7) - 4 = 31^\circ \), and checking the other expression, \( 3(7) + 10 = 31^\circ \). They agree. By the angle addition postulate, \( m\angle ABC = 31 + 31 = 62^\circ \). The question asked for the whole angle, not for \( x \) or for one part, which is the step most often skipped. \( 62^\circ \)

8. Two angles share vertex \( M \). A student refers to "\( \angle M \)". What should they write instead, and why?
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   The single-letter name is only unambiguous when exactly one angle has that vertex. With two angles at \( M \), the name \( \angle M \) could mean either, so a reader cannot tell which is being claimed about. They should use the three-letter name, with \( M \) in the middle and a point from each side on the outside: \( \angle PMQ \) and \( \angle QMR \), for instance. This matters more in geometry than it looks. A proof is an argument a reader must be able to check line by line, and a line that could mean two things cannot be checked. Ambiguous naming is marked wrong even when the intended meaning is guessable. A three-letter name with \( M \) in the middle, because two angles share that vertex

9. \( m\angle ABD = 2x \), \( m\angle DBC = 3x \), and \( \angle ABC \) is a right angle with \( D \) interior. Find both parts.
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   A right angle measures \( 90^\circ \) by definition, so that is the whole. By the angle addition postulate, \( 2x + 3x = 90 \), so \( 5x = 90 \) and \( x = 18 \). \( m\angle ABD = 2(18) = 36^\circ \) and \( m\angle DBC = 3(18) = 54^\circ \). Check: \( 36 + 54 = 90 \). Correct, and both are acute, which is necessarily true when two positive angles sum to a right angle. \( 36^\circ \) and \( 54^\circ \)

10. An angle measures \( (7x - 12)^\circ \) and its bisector creates two angles each measuring \( (2x + 9)^\circ \). Find the angle, and check the answer three ways.
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    The two halves must sum to the whole, by the angle addition postulate: \( (2x + 9) + (2x + 9) = 7x - 12 \), so \( 4x + 18 = 7x - 12 \), giving \( 30 = 3x \) and \( x = 10 \). The whole angle measures \( 7(10) - 12 = 58^\circ \), and each half measures \( 2(10) + 9 = 29^\circ \). Check one, addition: \( 29 + 29 = 58 \). Correct. Check two, halving: \( 58 \div 2 = 29 \), which is what a bisector must produce. Check three, classification: \( 58^\circ \) is acute and each half at \( 29^\circ \) is acute, which is consistent; bisecting an acute angle cannot produce anything but acute angles. \( 58^\circ \), with halves of \( 29^\circ \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the complement of \( 37^\circ \).
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   \( 90 - 37 \). \( 53^\circ \)

2. Find the supplement of \( 112^\circ \).
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   \( 180 - 112 \). \( 68^\circ \)

3. Two angles form a linear pair and one measures \( 43^\circ \). Find the other.
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   A linear pair is supplementary. \( 137^\circ \)

4. Two vertical angles measure \( (4x)^\circ \) and \( (2x + 30)^\circ \). Find \( x \).
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   Vertical angles are congruent: \( 4x = 2x + 30 \). \( x = 15 \)

5. Does an obtuse angle have a complement?
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   Its complement would have to be negative. No

6. An angle is \( 20^\circ \) more than its complement. Find both.
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   Let the smaller angle be \( x \), so the larger is \( x + 20 \). Complementary means they sum to 90: \( x + (x + 20) = 90 \), so \( 2x = 70 \) and \( x = 35 \). The angles are \( 35^\circ \) and \( 55^\circ \). Check: they sum to 90, and 55 is 20 more than 35. Both conditions hold. \( 35^\circ \) and \( 55^\circ \)

7. An angle is three times its supplement. Find both.
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   Let the smaller be \( x \), so the larger is \( 3x \). Supplementary means they sum to 180: \( x + 3x = 180 \), so \( 4x = 180 \) and \( x = 45 \). The angles are \( 45^\circ \) and \( 135^\circ \). Check: \( 45 + 135 = 180 \), and \( 135 = 3 \times 45 \). Both conditions hold. \( 45^\circ \) and \( 135^\circ \)

8. Give an example of two supplementary angles that do not form a linear pair.
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   Supplementary is defined by measure alone, so the angles need not touch at all. Take an angle of \( 60^\circ \) in one corner of a page and an angle of \( 120^\circ \) in the opposite corner. Their measures sum to \( 180^\circ \), so they are supplementary, and they share no vertex, so they cannot form a linear pair. A more useful example from later in the course: in a parallelogram, two consecutive angles are supplementary but the pair is not a linear pair, since their non-shared sides are two sides of the figure rather than one straight line. The general principle: a linear pair requires adjacency and a straight line, while supplementary requires only that the measures sum to 180. Linear pair is the stronger condition, and a proof that needs adjacency must establish it separately. Any two angles summing to \( 180^\circ \) that do not share a vertex and a side, such as \( 60^\circ \) and \( 120^\circ \) drawn separately

9. Two lines intersect. One angle measures \( 68^\circ \). Find all four angles.
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   Call the given angle \( \angle 1 \), with \( \angle 2 \) and \( \angle 4 \) adjacent to it and \( \angle 3 \) opposite. \( \angle 1 \) and \( \angle 2 \) form a linear pair, so they are supplementary: \( m\angle 2 = 180 - 68 = 112^\circ \). \( \angle 3 \) is vertical to \( \angle 1 \), so it is congruent: \( m\angle 3 = 68^\circ \). \( \angle 4 \) is vertical to \( \angle 2 \), so \( m\angle 4 = 112^\circ \). Check: \( 68 + 112 + 68 + 112 = 360^\circ \), a complete turn. \( 68^\circ \), \( 112^\circ \), \( 68^\circ \), \( 112^\circ \)

10. Two angles are both supplementary and congruent. Find their measures, and prove the result in general.
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    Let the measures be \( a \) and \( b \). Supplementary gives \( a + b = 180 \). Congruent gives \( a = b \) by the definition of congruent angles. Substituting: \( a + a = 180 \), so \( 2a = 180 \) and \( a = 90 \). Then \( b = 90 \) as well. So both angles are right angles. This is a small theorem worth remembering: if two angles are both congruent and supplementary, each is a right angle. It is the reason that two intersecting lines forming one right angle form four of them, which lesson 1.5 uses to define perpendicularity from a single right angle. The proof used both given facts and nothing from a diagram, which is what makes it general rather than a statement about one picture. Each measures \( 90^\circ \); two congruent supplementary angles are always right angles

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does \( \perp \) mean?
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   Perpendicular: the lines meet at a right angle

2. Two perpendicular lines meet. How many right angles are formed?
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   Four

3. A line passes through the midpoint of a segment. Is it the perpendicular bisector?
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   Only one condition is met. Not necessarily; it must also be perpendicular

4. Find the slope perpendicular to a line of slope \( \frac{2}{5} \).
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   \( -\frac{5}{2} \)

5. What reason justifies writing \( \angle ABD \cong \angle DBC \) when \( \overrightarrow{BD} \) bisects \( \angle ABC \)?
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   The definition of an angle bisector

6. Find the perpendicular bisector of the segment from \( (1, 2) \) to \( (7, 6) \).
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   Midpoint: \( \left( \dfrac{1+7}{2}, \dfrac{2+6}{2} \right) = (4, 4) \). Slope of the segment: \( \dfrac{6-2}{7-1} = \dfrac{4}{6} = \dfrac{2}{3} \). Perpendicular slope: \( -\dfrac{3}{2} \). Equation: \( y - 4 = -\dfrac{3}{2}(x - 4) \). Check that it passes through the midpoint: at \( x = 4 \) the right side is 0, so \( y = 4 \). Correct. \( y - 4 = -\frac{3}{2}(x - 4) \)

7. Verify that \( (7, 4) \) is equidistant from \( (1, 2) \) and \( (7, 6) \), and say what that suggests.
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   To \( (1,2) \): differences 6 and 2, so \( \sqrt{36 + 4} = \sqrt{40} \). To \( (7,6) \): differences 0 and 2, so \( \sqrt{0 + 4} = 2 \). These are not equal, since \( \sqrt{40} \approx 6.3 \). So \( (7,4) \) is not equidistant. Checking against the previous problem's answer: substituting \( x = 7 \) into \( y - 4 = -\frac{3}{2}(x - 4) \) gives \( y - 4 = -4.5 \), so \( y = -0.5 \), not 4. The point is not on the perpendicular bisector, which is consistent with the distances being unequal. The two facts always agree, which is the content of the perpendicular bisector theorem: a point is equidistant from the endpoints exactly when it lies on the perpendicular bisector. It is not equidistant, and correspondingly it does not lie on the perpendicular bisector

8. Explain why a definition can be used in both directions in a proof but a theorem usually cannot.
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   A definition states what a word means, so it is a biconditional: the term applies exactly when the condition holds. "A midpoint is a point that divides a segment into two congruent segments" means both that every midpoint does this and that anything doing this is a midpoint. A proof may therefore use it either way, citing "definition of midpoint" in each case. A theorem is a one-directional claim: if this, then that. Proving it establishes only that direction. "If two angles form a linear pair, then they are supplementary" does not by itself license concluding a linear pair from supplementarity, and in that case the converse is genuinely false, since supplementary angles need not be adjacent. A converse has to be proved separately, and some converses are true while others are not. Treating a theorem as reversible is one of the four errors this course names, and unit 3 is largely about keeping each theorem apart from its converse. A definition is a biconditional; a theorem states one direction, and its converse needs its own proof

9. \( \overrightarrow{BD} \) bisects \( \angle ABC \). \( m\angle ABD = (4x - 7)^\circ \) and \( m\angle DBC = (2x + 9)^\circ \). Is \( \angle ABC \) acute, right or obtuse?
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   By the definition of an angle bisector the two parts are congruent, so their measures are equal: \( 4x - 7 = 2x + 9 \), giving \( 2x = 16 \) and \( x = 8 \). Each part measures \( 4(8) - 7 = 25^\circ \), confirmed by the other expression: \( 2(8) + 9 = 25^\circ \). By the angle addition postulate, \( m\angle ABC = 25 + 25 = 50^\circ \). Since \( 50 \lt 90 \), the angle is acute. Acute, measuring \( 50^\circ \)

10. A line \( \ell \) is perpendicular to \( \overline{AB} \) and intersects it at a point \( P \) with \( AP = 5 \) and \( PB = 5 \). Prove \( \ell \) is the perpendicular bisector, naming every reason.
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    Given: \( \ell \perp \overline{AB} \), \( \ell \) meets \( \overline{AB} \) at \( P \), \( AP = 5 \), \( PB = 5 \). Prove: \( \ell \) is the perpendicular bisector of \( \overline{AB} \). 1. \( AP = 5 \) and \( PB = 5 \). Reason: given. 2. \( AP = PB \). Reason: transitive property of equality, since both equal 5. 3. \( \overline{AP} \cong \overline{PB} \). Reason: definition of congruent segments. 4. \( P \) lies on \( \overline{AB} \) between \( A \) and \( B \). Reason: given, since \( \ell \) intersects the segment at \( P \). 5. \( P \) is the midpoint of \( \overline{AB} \). Reason: definition of midpoint, using steps 3 and 4. 6. \( \ell \) passes through the midpoint of \( \overline{AB} \). Reason: step 5 and the given that \( \ell \) meets the segment at \( P \). 7. \( \ell \perp \overline{AB} \). Reason: given. 8. \( \ell \) is the perpendicular bisector of \( \overline{AB} \). Reason: definition of perpendicular bisector, using steps 6 and 7. Both halves of the definition were established separately and then combined, which is exactly what the definition demands. Step 4 is the one students omit: without betweenness, a point equidistant from \( A \) and \( B \) need not be on the segment at all. Proved, with both conditions of the definition established

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What two tools are allowed in a construction?
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   A compass and an unmarked straightedge

2. Why is the straightedge unmarked?
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   So it cannot be used to measure, only to draw a line through two points

3. In bisecting a segment, must the compass width exceed half the length?
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   Otherwise the arcs never meet. Yes

4. What does a compass guarantee that a ruler does not?
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   That two distances are exactly equal, without measuring either

5. Describe the first step in bisecting an angle.
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   Place the compass point on the vertex and draw an arc crossing both sides

6. Describe the full construction for copying an angle.
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   Given \( \angle A \) and a ray with endpoint \( D \). Place the compass point on \( A \) and draw an arc crossing both sides of the angle at points \( B \) and \( C \). Without changing the width, place the compass point on \( D \) and draw a large arc crossing the new ray at a point \( E \). Set the compass to the width \( BC \), the distance between the two crossing points of the first arc. Place the compass point on \( E \) and draw an arc crossing the large arc at a point \( F \). Draw \( \overrightarrow{DF} \). Then \( \angle FDE \cong \angle A \). It works because \( \triangle ABC \) and \( \triangle DEF \) have three pairs of congruent sides by construction, so they are congruent by SSS, and corresponding angles of congruent triangles are congruent. Arc across both sides, copy that arc at the new vertex, transfer the chord width, draw the ray

7. Describe how to construct a perpendicular to a line through a point on that line.
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   Let the point be \( P \) on line \( \ell \). Place the compass point on \( P \) and draw arcs crossing \( \ell \) on both sides, at points \( A \) and \( B \). By construction \( PA = PB \), so \( P \) is now the midpoint of \( \overline{AB} \). Now construct the perpendicular bisector of \( \overline{AB} \): widen the compass, draw arcs from \( A \) and from \( B \) that cross above the line, and draw the line from that crossing point through \( P \). That line is perpendicular to \( \ell \) and passes through \( P \), as required. The construction reduces to the one already proved, which is the usual pattern: later constructions are built from earlier ones rather than from scratch. Mark equal distances either side of \( P \), then bisect that segment

8. Explain why a construction is closer to a proof than to a drawing.
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   A drawing is as accurate as the hand that made it and the instrument that measured it. Measuring an angle with a protractor and drawing its bisector produces something approximately right, limited by the protractor's precision and by eyesight. A construction makes no measurements at all. It produces equalities by the compass, which guarantees that two distances are the same because the opening never changed, and then it derives the result from those equalities. The output is exactly correct as a matter of logic, not approximately correct as a matter of care. That is why every construction in this course is followed by its proof. The construction is the claim and the proof is the justification, exactly as elsewhere in geometry. A student who can perform the steps but cannot say why they work has learned the drawing and not the mathematics. It produces an exact result guaranteed by an argument, rather than an approximate one limited by measurement

9. A student bisects a segment but changes the compass width between the arcs from \( A \) and the arcs from \( B \). What goes wrong?
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   The whole proof rests on the crossing points being equidistant from \( A \) and from \( B \). If the width changes, the crossing point \( P \) satisfies \( PA = r_1 \) and \( PB = r_2 \) with \( r_1 \neq r_2 \), so \( PA \neq PB \). Without that equality, \( \triangle PAQ \) and \( \triangle PBQ \) are not congruent, nothing follows about the angles, and the line through the crossing points is neither perpendicular to \( \overline{AB} \) nor through its midpoint. Geometrically the line will still cross the segment, but at the wrong place and the wrong angle, and it will look close enough to pass a glance. That is what makes the error dangerous: the failure is invisible in the drawing and only the argument reveals it. The crossing points are no longer equidistant from both endpoints, so the line is neither perpendicular nor a bisector

10. Describe how to construct an equilateral triangle on a given segment, and prove it is equilateral.
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    Given \( \overline{AB} \). Set the compass to the width \( AB \). Place the point on \( A \) and draw an arc above the segment. Without changing the width, place the point on \( B \) and draw a second arc crossing the first at a point \( C \). Draw \( \overline{AC} \) and \( \overline{BC} \). Proof. The compass was set to \( AB \) and never changed. \( C \) lies on the arc centered at \( A \) with radius \( AB \), so \( AC = AB \). \( C \) also lies on the arc centered at \( B \) with the same radius, so \( BC = AB \). By the transitive property, \( AC = BC \), and all three sides have length \( AB \). A triangle with three congruent sides is equilateral by definition, so \( \triangle ABC \) is equilateral. This is the first proposition in Euclid's Elements, and it is a good example of how little machinery a construction proof needs: one fixed compass width and the transitive property. Two arcs of radius \( AB \) centered at \( A \) and \( B \); all three sides equal \( AB \) because the compass never changed

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Name a polygon with seven sides.
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   Heptagon

2. What makes a polygon regular?
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   All sides congruent and all angles congruent

3. Is a circle a polygon?
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   It is not made of segments. No

4. How many diagonals does a pentagon have?
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   \( \dfrac{5(2)}{2} = 5 \). 5

5. What does concave mean?
   Show the full solution

   At least one interior angle exceeds \( 180^\circ \)

6. How many diagonals does a decagon have?
   Show the full solution

   \( \dfrac{n(n-3)}{2} \) with \( n = 10 \): \( \dfrac{10(7)}{2} = 35 \). Checking the reasoning directly: from each of the 10 vertices, diagonals go to \( 10 - 3 = 7 \) others, giving 70 endpoints, and each diagonal has two endpoints, so \( 70 \div 2 = 35 \). 35

7. A polygon has 27 diagonals. How many sides does it have?
   Show the full solution

   Set the formula equal to 27: \( \dfrac{n(n-3)}{2} = 27 \), so \( n(n - 3) = 54 \) and \( n^2 - 3n - 54 = 0 \). Factoring: two numbers multiplying to \( -54 \) and adding to \( -3 \) are \( -9 \) and 6, so \( (n - 9)(n + 6) = 0 \), giving \( n = 9 \) or \( n = -6 \). A polygon cannot have a negative number of sides, so \( n = 9 \). Check: \( \dfrac{9(6)}{2} = 27 \). Correct. The figure is a nonagon. 9 sides

8. Is an equilateral polygon necessarily regular? Is an equiangular one?
   Show the full solution

   Neither implication holds for polygons in general, and a counterexample settles each. Equilateral but not regular: a rhombus that is not a square. All four sides are congruent, but the angles come in two different sizes, so it is not equiangular. Equiangular but not regular: a rectangle that is not a square. All four angles are right angles, but the sides come in two different lengths. So both conditions are genuinely needed in the definition of regular. The triangle is the exception worth knowing: an equilateral triangle is necessarily equiangular and the reverse also holds, which lesson 5.7 proves. Triangles are rigid in a way other polygons are not, and that rigidity is why the triangle congruence criteria of unit 5 exist at all. No to both; a rhombus and a rectangle are the counterexamples, though triangles are a special case where each does imply the other

9. Explain why the definition of a polygon requires the sides to meet only at their endpoints.
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   Without that condition, a figure whose sides cross in the middle would count as a polygon, and almost every theorem in the course would become false. Consider a four-segment closed path where two sides cross, sometimes drawn as a bowtie. Its interior is not well defined, so "interior angle" has no clear meaning and the angle sum formula of lesson 7.1 does not apply. Its area cannot be computed by the usual methods because the two loops would count with opposite orientation. The condition is doing real work: it guarantees that the figure separates the plane into a well-defined inside and outside, which is what makes interior angles, area, and the whole apparatus of unit 7 and unit 11 meaningful. This is a general feature of geometric definitions. Each clause excludes a specific pathological case, and reading a definition carefully means asking what each clause is there to rule out. It guarantees a well-defined interior, without which interior angles and area would be meaningless

10. A regular polygon has a diagonal count equal to twice its number of sides. Find the polygon.
    Show the full solution

    Let \( n \) be the number of sides. The condition is \( \dfrac{n(n-3)}{2} = 2n \). Multiply both sides by 2: \( n(n - 3) = 4n \), so \( n^2 - 3n = 4n \) and \( n^2 - 7n = 0 \). Factor: \( n(n - 7) = 0 \), giving \( n = 0 \) or \( n = 7 \). A polygon must have at least three sides, so \( n = 0 \) is rejected on those grounds and \( n = 7 \). Check: a heptagon has \( \dfrac{7(4)}{2} = 14 \) diagonals, and twice its number of sides is \( 2(7) = 14 \). They match. Note that the factoring had to avoid dividing both sides by \( n \), which would have discarded the root \( n = 0 \). It happens to be a root that gets rejected anyway, but relying on that is luck rather than method. A heptagon, with 7 sides and 14 diagonals

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is inductive reasoning?
   Show the full solution

   Forming a general conjecture from specific cases

2. How many counterexamples are needed to disprove a conjecture?
   Show the full solution

   One

3. Predict the next term: 2, 6, 12, 20, 30.
   Show the full solution

   Differences are 4, 6, 8, 10, so the next difference is 12. 42

4. Find a counterexample to "all quadrilaterals have four congruent sides."
   Show the full solution

   Any rectangle that is not a square

5. Does checking ten cases prove a conjecture?
   Show the full solution

   No; it only supports it

6. Find a counterexample to "if a number is divisible by 3, then it is odd."
   Show the full solution

   A counterexample must satisfy the hypothesis and fail the conclusion, so it must be a multiple of 3 that is even. Take 6: it is divisible by 3, and it is even, not odd. The claim fails. Note that 4 would not work as a counterexample, because it fails the hypothesis: it is not divisible by 3, so the statement says nothing about it. 6, or any even multiple of 3

7. A student conjectures that \( n^2 + n + 11 \) is prime for every whole number \( n \), having checked \( n = 0 \) through \( n = 9 \). Evaluate the conjecture.
   Show the full solution

   The checks do hold: at \( n = 0 \) it gives 11, at \( n = 1 \) it gives 13, and so on up to \( n = 9 \) giving \( 81 + 9 + 11 = 101 \), all prime. Ten confirming cases is unusually strong evidence. It is still not a proof, and the conjecture is false. At \( n = 11 \): \( 121 + 11 + 11 = 143 \), and \( 143 = 11 \times 13 \), so it is not prime. The failure is not accidental. Whenever \( n \) is a multiple of 11, every term is divisible by 11, so the whole expression is. At \( n = 11 \) that is the first case to occur, which is exactly why ten checks missed it. This is the lesson's point in its sharpest form: a pattern can hold for ten consecutive cases and fail on the eleventh, and no amount of checking distinguishes a true conjecture from one that fails later. False; \( n = 11 \) gives \( 143 = 11 \times 13 \)

8. Explain why a counterexample must satisfy the hypothesis of the conditional.
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   A conditional statement claims only that whenever the hypothesis holds, the conclusion follows. It makes no claim at all about cases where the hypothesis fails. Take "if a figure is a square, then it has four right angles." Offering a triangle as a counterexample proves nothing, because the statement never said anything about triangles. The triangle is not a case the claim covers. A genuine counterexample has to be a case the claim does cover, where the promised conclusion nevertheless fails. Here that would require a square without four right angles, which does not exist, and correspondingly the statement is true. The practical rule: to disprove "if p then q", find something that is p and is not q. Anything that is not p is irrelevant. A conditional claims nothing about cases failing its hypothesis, so such a case cannot contradict it

9. Make a conjecture about the sum of the first \( n \) odd numbers, then prove it geometrically.
   Show the full solution

   Compute the first few sums. \( 1 = 1 \). \( 1 + 3 = 4 \). \( 1 + 3 + 5 = 9 \). \( 1 + 3 + 5 + 7 = 16 \). \( 1 + 3 + 5 + 7 + 9 = 25 \). The sums are \( 1, 4, 9, 16, 25 \), which are \( 1^2, 2^2, 3^2, 4^2, 5^2 \). Conjecture: the sum of the first \( n \) odd numbers is \( n^2 \). Geometric proof. Build a square of dots. Start with one dot, a \( 1 \times 1 \) square. To turn an \( n \times n \) square into an \( (n+1) \times (n+1) \) square, add a column of \( n \) dots on the right, a row of \( n \) dots on top, and one corner dot: that is \( 2n + 1 \) dots, which is an odd number, and it is the next odd number each time. So the \( n \times n \) square is built from one dot plus successive odd numbers of dots, and it contains \( n^2 \) dots. That is the conjecture, proved for every \( n \) at once rather than checked for a few. The sum is \( n^2 \), proved by building a square one L-shaped layer at a time

10. A student says "I found a counterexample, so the conjecture is probably false." Correct them.
    Show the full solution

    "Probably" understates it. A counterexample does not make a conjecture unlikely; it makes it false, definitively and permanently. A general claim says something holds in every case. One case where it does not hold contradicts it outright. There is nothing left to decide and no further evidence that could rescue it. The asymmetry is worth stating plainly, because it runs opposite to how evidence works in most subjects. Confirming cases are weak: any number of them leaves the claim unproved. A disconfirming case is decisive: one of them settles the matter. What may remain is a repaired conjecture. If "\( n^2 \gt n \) for every number" fails at \( n = 1 \) and on the interval from 0 to 1, the salvageable version is "\( n^2 \gt n \) for every \( n \gt 1 \)", which is true and provable. Finding a counterexample often tells you what the correct hypothesis should have been. A counterexample makes the conjecture false, not merely improbable, though it may point to a corrected version

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Identify the hypothesis of "if it rains, then the ground is wet."
   Show the full solution

   It rains

2. Write the converse of "if \( x = 3 \), then \( x^2 = 9 \)."
   Show the full solution

   If \( x^2 = 9 \), then \( x = 3 \)

3. Is that converse true?
   Show the full solution

   \( x = -3 \) gives \( x^2 = 9 \). No

4. Which statement is always equivalent to a conditional?
   Show the full solution

   Its contrapositive

5. Rewrite "all squares are rectangles" in if-then form.
   Show the full solution

   If a figure is a square, then it is a rectangle

6. Write all three related statements for "if a figure is a square, then it has four right angles," and evaluate each.
   Show the full solution

   Original: if a figure is a square, then it has four right angles. True, by the definition of a square. Converse: if a figure has four right angles, then it is a square. False; a non-square rectangle has four right angles. Inverse: if a figure is not a square, then it does not have four right angles. False; the same rectangle is the counterexample, as it must be since the inverse is equivalent to the converse. Contrapositive: if a figure does not have four right angles, then it is not a square. True, as it must be since it is equivalent to the original. Original and contrapositive true; converse and inverse false

7. Write the contrapositive of "if two lines are parallel, then they do not intersect," and say whether it is true.
   Show the full solution

   Swap and negate both parts. The conclusion "they do not intersect" negates to "they do intersect," and the hypothesis "they are parallel" negates to "they are not parallel." Contrapositive: if two lines intersect, then they are not parallel. It is true, and it must be, since the original is true and a statement is always equivalent to its contrapositive. This one is worth noticing because the contrapositive is arguably the more useful form: it is what licenses concluding non-parallelism from an intersection, which comes up throughout unit 3. If two lines intersect, then they are not parallel; true

8. Explain why a definition can be written as a biconditional but most theorems cannot.
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   A definition assigns a name to a condition, so the name applies exactly when the condition holds. "A right angle is an angle measuring \( 90^\circ \)" means both that every right angle measures \( 90^\circ \) and that every angle measuring \( 90^\circ \) is a right angle. Both directions are built into what a definition is, so it can be written "an angle is right if and only if it measures \( 90^\circ \)." A theorem asserts one direction, and whether the other direction also holds is a separate question with its own answer. "If two angles are vertical, then they are congruent" is true; its converse is false. Nothing about proving the first tells you anything about the second. The practical consequence in proofs: a definition may be cited in either direction with the same reason, while a theorem may only be cited in the direction that was proved. Its converse, if true, is a different theorem with a different name. A definition asserts both directions by its nature; a theorem asserts one, and its converse requires separate proof

9. Give a statement whose converse is also true, and one whose converse is false.
   Show the full solution

   Converse true: "If a triangle is equilateral, then it is equiangular." Its converse, "if a triangle is equiangular, then it is equilateral," is also true, proved in lesson 5.7. Because both directions hold, the two conditions can be written as a biconditional: a triangle is equilateral if and only if it is equiangular. Converse false: "If two angles form a linear pair, then they are supplementary." Its converse, "if two angles are supplementary, then they form a linear pair," is false: two angles of \( 60^\circ \) and \( 120^\circ \) drawn in different places sum to \( 180^\circ \) without being adjacent. The pair illustrates the whole point of the lesson. From the outside these look like the same kind of statement, and only proving or disproving each converse separately settles which is which. Equilateral and equiangular triangles: converse true. Linear pair and supplementary: converse false

10. A proof cites "converse of the alternate interior angles theorem." Explain what that means and why the name matters.
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    The alternate interior angles theorem, proved in lesson 3.2, says: if two parallel lines are cut by a transversal, then the alternate interior angles are congruent. It takes parallelism as given and produces congruent angles. Its converse, proved separately in lesson 3.3, says: if two lines cut by a transversal form congruent alternate interior angles, then the lines are parallel. It takes congruent angles as given and produces parallelism. A proof citing the converse is therefore doing the opposite job: concluding that lines are parallel rather than using that they are. Naming which direction is being used tells the reader what the line of the proof is doing, and it is the check that prevents circular reasoning, since a proof cannot assume parallelism in order to establish it. This is why unit 3 names each theorem and each converse separately rather than treating them as one fact. Both happen to be true here, which is exactly what makes the confusion easy: a student who assumes theorems are reversible will be right in unit 3 and wrong elsewhere, and will not know the difference. It cites the parallel-concluding direction rather than the parallel-assuming one, and naming it keeps the proof from being circular

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does the law of detachment let you conclude?
   Show the full solution

   The conclusion, when the conditional is true and its hypothesis holds

2. What does the law of syllogism produce?
   Show the full solution

   A new conditional linking the first hypothesis to the last conclusion

3. Is deductive reasoning certain or probable?
   Show the full solution

   Certain, given true premises and a valid form

4. "If \( x \gt 5 \), then \( x \gt 2 \). And \( x = 7 \)." What follows?
   Show the full solution

   \( 7 \gt 5 \), so the hypothesis holds and detachment applies. \( x \gt 2 \)

5. "If \( x \gt 5 \), then \( x \gt 2 \). And \( x = 3 \)." What follows about \( x \gt 5 \)?
   Show the full solution

   \( 3 \gt 2 \) is true but \( 3 \gt 5 \) is false, so the conclusion holding tells you nothing. Nothing; concluding \( x \gt 5 \) would be affirming the conclusion

6. Chain these: if a polygon is regular, all its sides are congruent. If all a polygon's sides are congruent, its perimeter is the side length times the number of sides.
   Show the full solution

   The conclusion of the first, "all its sides are congruent," is exactly the hypothesis of the second, so the law of syllogism applies. If a polygon is regular, then its perimeter is the side length times the number of sides

7. Is this valid? "If two lines are perpendicular, they form a right angle. Lines \( a \) and \( b \) form a right angle. Therefore \( a \perp b \)."
   Show the full solution

   The form is affirming the conclusion, which is invalid in general: the second premise gives the conclusion of the conditional, and the argument infers the hypothesis. However, the conclusion happens to be true, because perpendicularity is a definition rather than a theorem. "Two lines are perpendicular if and only if they form a right angle" is a biconditional, so the reverse direction is separately available and may be cited. The argument as written is defective even though its conclusion is correct. The right justification is "definition of perpendicular lines" used in the reverse direction, not the law of detachment applied to a conditional. This distinction is worth the care: the same reasoning applied to a theorem instead of a definition would produce a false conclusion. The form is invalid, but the conclusion is justified because perpendicularity is a definition and therefore reversible

8. Explain the difference between the law of detachment and the law of syllogism.
   Show the full solution

   They take different inputs and produce different kinds of output. Detachment takes one conditional and one specific fact, and produces a specific fact. From "if an angle is right, it measures \( 90^\circ \)" plus "\( \angle A \) is right," it produces "\( m\angle A = 90 \)." The output is about a particular object. Syllogism takes two conditionals and produces a third conditional. From "if regular, then equilateral" plus "if equilateral, then all sides congruent," it produces "if regular, then all sides congruent." The output is another general rule, and no particular figure has been mentioned. In a two-column proof, detachment is what almost every line does: a theorem is cited and applied to the figure at hand. Syllogism appears more often in the background, when a chain of theorems is compressed into one step or when a new theorem is derived from existing ones. Detachment applies a rule to a case and yields a fact; syllogism chains two rules and yields a new rule

9. Construct an invalid argument that looks convincing, and identify the error.
   Show the full solution

   The argument: If a quadrilateral is a rhombus, then its diagonals are perpendicular. Quadrilateral \( ABCD \) has perpendicular diagonals. Therefore \( ABCD \) is a rhombus. Why it looks convincing: the conditional is a genuine theorem from lesson 7.4, the premise is a real geometric fact, and rhombuses do have perpendicular diagonals, so the conclusion feels natural. The error: affirming the conclusion. The argument reasons backward along a theorem, which is the converse error. The counterexample: a kite that is not a rhombus has perpendicular diagonals, as lesson 7.5 shows. So the premises hold and the conclusion fails. The converse does become true with an extra condition: a quadrilateral whose diagonals are perpendicular and bisect each other is a rhombus. Finding what the converse needs in order to be true is often more useful than simply noting that it fails. Affirming the conclusion; a non-rhombus kite is the counterexample

10. Given: if a triangle is equilateral, it is isosceles. If a triangle is isosceles, it has two congruent angles. \( \triangle ABC \) has no two congruent angles. What follows?
    Show the full solution

    Two moves are needed, and the order matters. First, syllogism. The conclusion of the first conditional is the hypothesis of the second, so they chain: if a triangle is equilateral, then it has two congruent angles. Second, the contrapositive. The given fact denies the conclusion of that chained conditional: \( \triangle ABC \) does not have two congruent angles. Denying the conclusion is the valid direction, so it follows that \( \triangle ABC \) is not equilateral. What also follows. Applying the contrapositive of the second conditional alone gives that \( \triangle ABC \) is not isosceles. So the triangle is scalene. What does not follow. Nothing about whether it is right, acute or obtuse; the premises say nothing about that. \( \triangle ABC \) is neither equilateral nor isosceles, so it is scalene

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What property justifies going from \( x - 4 = 9 \) to \( x = 13 \)?
   Show the full solution

   Addition property of equality

2. What property justifies going from \( 5x = 30 \) to \( x = 6 \)?
   Show the full solution

   Division property of equality

3. What property is \( \overline{AB} \cong \overline{AB} \)?
   Show the full solution

   Reflexive property of congruence

4. What property justifies going from \( 2(x + 3) \) to \( 2x + 6 \)?
   Show the full solution

   Distributive property

5. If \( AB = CD \) and \( CD = EF \), what property gives \( AB = EF \)?
   Show the full solution

   Transitive property of equality

6. Prove that if \( 4x + 7 = 31 \), then \( x = 6 \), with a reason for every line.
   Show the full solution

   Given: \( 4x + 7 = 31 \). Prove: \( x = 6 \). 1. \( 4x + 7 = 31 \). Reason: given. 2. \( 4x = 24 \). Reason: subtraction property of equality. 3. \( x = 6 \). Reason: division property of equality. Check: \( 4(6) + 7 = 31 \). Correct. Proved in three lines

7. Prove that if \( \dfrac{x}{3} - 2 = 7 \), then \( x = 27 \).
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   Given: \( \dfrac{x}{3} - 2 = 7 \). Prove: \( x = 27 \). 1. \( \dfrac{x}{3} - 2 = 7 \). Reason: given. 2. \( \dfrac{x}{3} = 9 \). Reason: addition property of equality. 3. \( x = 27 \). Reason: multiplication property of equality, multiplying both sides by 3. Check: \( 27 \div 3 - 2 = 9 - 2 = 7 \). Correct. Proved in three lines

8. Prove that if \( 5(y - 2) = 3y + 8 \), then \( y = 9 \).
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   Given: \( 5(y - 2) = 3y + 8 \). Prove: \( y = 9 \). 1. \( 5(y - 2) = 3y + 8 \). Reason: given. 2. \( 5y - 10 = 3y + 8 \). Reason: distributive property. 3. \( 2y - 10 = 8 \). Reason: subtraction property of equality, subtracting \( 3y \). 4. \( 2y = 18 \). Reason: addition property of equality, adding 10. 5. \( y = 9 \). Reason: division property of equality. Check: \( 5(9 - 2) = 35 \) and \( 3(9) + 8 = 35 \). They agree. Proved in five lines

9. Explain why an algebraic proof is worth writing when you already know how to solve the equation.
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   The point is not the answer, which is already reachable, but the habit of justifying each move by name. Geometry will shortly ask for arguments about figures, where there is no procedure to fall back on and where the reason for each step is the entire content of the work. A student who has never articulated why subtracting from both sides is allowed will struggle to articulate why two triangles are congruent, because the difficulty is the same and only the subject matter differs. There is a second reason. The properties of equality used here are the same properties used in geometric proofs, extended to congruence. The reflexive property is what lets a shared side enter a triangle congruence proof. The substitution property is what lets a length established in one part of a figure be used in another. Practicing them on equations, where the mathematics is easy, means only one new thing has to be learned at a time. It builds the habit of naming reasons, on familiar material, using the same properties geometric proofs will need

10. Prove that if \( M \) is the midpoint of \( \overline{AB} \), then \( AM = \frac{1}{2}AB \), naming every reason.
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    Given: \( M \) is the midpoint of \( \overline{AB} \). Prove: \( AM = \frac{1}{2}AB \). 1. \( M \) is the midpoint of \( \overline{AB} \). Reason: given. 2. \( \overline{AM} \cong \overline{MB} \). Reason: definition of midpoint. 3. \( AM = MB \). Reason: definition of congruent segments. 4. \( AM + MB = AB \). Reason: segment addition postulate, since a midpoint lies between the endpoints. 5. \( AM + AM = AB \). Reason: substitution property of equality, replacing \( MB \) with \( AM \) using line 3. 6. \( 2AM = AB \). Reason: distributive property. 7. \( AM = \frac{1}{2}AB \). Reason: division property of equality. This is the first proof in the course that mixes geometric reasons with algebraic ones, and it shows how they interlock. Lines 2 and 4 come from geometry; lines 5, 6 and 7 are pure algebra; line 3 is the bridge between the geometric language of congruence and the arithmetic language of length. That bridge line is the one students omit, and without it line 5 has nothing to substitute. Proved in seven lines, with the congruence converted to an equation before the algebra begins

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What are the two columns of a two-column proof?
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   Statements and reasons

2. What is the reason for the first line of a proof?
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   Given

3. Is "it looks that way in the diagram" a valid reason?
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   No

4. What must the last statement of a proof be?
   Show the full solution

   Exactly what was to be proved

5. What reason converts \( \angle A \cong \angle B \) into \( m\angle A = m\angle B \)?
   Show the full solution

   Definition of congruent angles

6. Given \( M \) is the midpoint of \( \overline{AB} \) and \( AB = 24 \), prove \( MB = 12 \).
   Show the full solution

   1. \( M \) is the midpoint of \( \overline{AB} \); \( AB = 24 \). Reason: given. 2. \( \overline{AM} \cong \overline{MB} \). Reason: definition of midpoint. 3. \( AM = MB \). Reason: definition of congruent segments. 4. \( AM + MB = AB \). Reason: segment addition postulate. 5. \( MB + MB = 24 \). Reason: substitution property of equality, using lines 1 and 3. 6. \( 2 \cdot MB = 24 \). Reason: distributive property. 7. \( MB = 12 \). Reason: division property of equality. Proved in seven lines

7. Given that \( \angle 1 \) and \( \angle 2 \) form a linear pair and \( m\angle 1 = 62^\circ \), prove \( m\angle 2 = 118^\circ \).
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   1. \( \angle 1 \) and \( \angle 2 \) form a linear pair; \( m\angle 1 = 62 \). Reason: given. 2. \( \angle 1 \) and \( \angle 2 \) are supplementary. Reason: linear pair theorem. 3. \( m\angle 1 + m\angle 2 = 180 \). Reason: definition of supplementary angles. 4. \( 62 + m\angle 2 = 180 \). Reason: substitution property of equality. 5. \( m\angle 2 = 118 \). Reason: subtraction property of equality. Line 2 and line 3 are separate for a reason: the linear pair theorem concludes supplementarity, and the definition of supplementary converts that into an equation. Collapsing them hides which fact is doing the work. Proved in five lines

8. A proof's reason column reads: given, given, obvious, so they are congruent. Diagnose it.
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   Two of the four entries are defective, and both defects are fatal. "Obvious" is not a reason. It names nothing that could be checked, and it is exactly what the format exists to prevent. Whatever step it justifies either follows from a definition, postulate or theorem, in which case that should be named, or it does not follow, in which case the proof has a gap. "So they are congruent" is a statement, not a reason. It belongs in the left column. The right column must say what licenses that conclusion, such as a congruence criterion or the definition of congruent segments. The test named in the method is what catches both: read only the reason column. It should be a list of specific, citable results. Anything that is vague, or that repeats the statement, or that appeals to appearance, is a gap in the argument rather than a stylistic shortfall. "Obvious" names no justification, and "so they are congruent" is a statement misplaced in the reason column

9. Given \( \angle A \cong \angle B \), \( \angle B \cong \angle C \), and \( m\angle C = 40^\circ \), prove \( m\angle A = 40^\circ \).
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   1. \( \angle A \cong \angle B \); \( \angle B \cong \angle C \); \( m\angle C = 40 \). Reason: given. 2. \( \angle A \cong \angle C \). Reason: transitive property of congruence. 3. \( m\angle A = m\angle C \). Reason: definition of congruent angles. 4. \( m\angle A = 40 \). Reason: substitution property of equality. An alternative route converts to measures first and then uses the transitive property of equality instead of congruence. Both are correct; what matters is that the property named matches the relation being used, congruence or equality, on that line. Proved in four lines

10. Explain why the two-column format is used to teach proof, and what it makes impossible.
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    The format enforces one rule by its shape: every statement sits beside a reason, so an unjustified claim leaves a visible blank. In a paragraph, the same gap can hide behind a connective. "Clearly, therefore, and so it follows that" reads smoothly while justifying nothing, and both the writer and the reader can miss it. It also makes the structure of the argument visible. The statements form a chain from the given to the conclusion, each line depending on earlier ones, and reading down the left column shows the path taken. A proof that jumps, or that assumes something not yet established, shows the jump as a line whose reason cannot be supplied. A third thing it prevents is circularity. Since every reason must be a result already available, citing the theorem being proved is immediately visible as a reason that is not yet legitimate. What the format costs is readability. Mathematicians write proofs in paragraphs because they are easier to follow once the habit of justification is secure. Lesson 2.7 takes that up. The two-column form is a training format, kept until the discipline it enforces no longer needs enforcing. It makes an unjustified assertion visible as an empty reason, and it exposes gaps and circular citations that a paragraph can hide

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the congruent complements theorem.
   Show the full solution

   If two angles are complementary to the same angle or to congruent angles, then they are congruent

2. Are all right angles congruent?
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   Each measures \( 90^\circ \). Yes

3. \( \angle 1 \) and \( \angle 2 \) are both supplementary to \( \angle 3 \). What follows?
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   \( \angle 1 \cong \angle 2 \), by the congruent supplements theorem

4. May a theorem be used inside its own proof?
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   No; that is circular reasoning

5. \( \angle A \) is complementary to \( \angle B \), and \( m\angle A = 28^\circ \). Find \( m\angle B \).
   Show the full solution

   \( 62^\circ \)

6. Prove the congruent complements theorem.
   Show the full solution

   Given: \( \angle 1 \) and \( \angle 2 \) are complementary; \( \angle 3 \) and \( \angle 2 \) are complementary. Prove: \( \angle 1 \cong \angle 3 \). 1. Both pairs are complementary. Reason: given. 2. \( m\angle 1 + m\angle 2 = 90 \). Reason: definition of complementary angles. 3. \( m\angle 3 + m\angle 2 = 90 \). Reason: definition of complementary angles. 4. \( m\angle 1 + m\angle 2 = m\angle 3 + m\angle 2 \). Reason: transitive property of equality. 5. \( m\angle 1 = m\angle 3 \). Reason: subtraction property of equality. 6. \( \angle 1 \cong \angle 3 \). Reason: definition of congruent angles. The proof is identical to the supplements version with 90 in place of 180, which is worth noticing: the argument never used anything about the number itself. Proved in six lines

7. \( \angle 1 \) and \( \angle 2 \) form a linear pair, and \( \angle 2 \) and \( \angle 3 \) form a linear pair. Prove \( \angle 1 \cong \angle 3 \).
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   1. \( \angle 1 \) and \( \angle 2 \) form a linear pair; \( \angle 2 \) and \( \angle 3 \) form a linear pair. Reason: given. 2. \( \angle 1 \) and \( \angle 2 \) are supplementary. Reason: linear pair theorem. 3. \( \angle 2 \) and \( \angle 3 \) are supplementary. Reason: linear pair theorem. 4. \( \angle 1 \cong \angle 3 \). Reason: congruent supplements theorem, since both are supplementary to \( \angle 2 \). This is exactly the vertical angles theorem, since two angles each forming a linear pair with a third are vertical to each other. Proving it this way instead of from scratch shows what a named theorem buys: the whole algebraic middle of the lesson 1.4 proof is now compressed into line 4. Proved in four lines using the congruent supplements theorem

8. Explain why the congruent supplements theorem does not require the angles to be adjacent.
   Show the full solution

   Look at what the proof actually used. It converted "supplementary" into an equation about measures, set the two equations equal because both sides equaled 180, and canceled the shared term. Every step was about numbers. Nothing in that chain referred to where the angles were, whether they touched, or whether they appeared in the same figure. Supplementary is defined by measure alone, as lesson 1.4 emphasized, and the proof inherits that generality. The practical consequence is useful. The theorem applies to two angles in completely different parts of a complicated diagram, provided each is supplementary to the same third angle. Proofs in unit 3 use it exactly that way, relating an angle above a transversal to one several intersections away. A theorem is only as general as its proof allows, and reading a proof to see what it did not use is how you find out how widely it applies. The proof used only the numerical definition of supplementary, so position never entered it

9. A student proves the vertical angles theorem by citing "vertical angles are congruent" at line 3. Diagnose it.
   Show the full solution

   This is circular reasoning, the second of the four errors this course names. The proof assumes exactly what it is supposed to establish. A proof must derive its conclusion from results already available, meaning things proved earlier or accepted as postulates. The vertical angles theorem is not yet available inside its own proof, because its availability is what the proof is creating. The failure is easy to miss because the cited statement is true. Circularity does not produce a false conclusion; it produces no conclusion at all, since the argument establishes nothing that was not already assumed. If the statement were unavailable for some other reason, the same proof would collapse. The correct proof, from lesson 1.4, uses two applications of the linear pair theorem and the subtraction property of equality, all of which were available before the theorem existed. Circular reasoning: the theorem is cited inside its own proof, so nothing is established

10. \( \angle 1 \) is supplementary to \( \angle 2 \), \( \angle 3 \) is supplementary to \( \angle 4 \), and \( \angle 2 \cong \angle 4 \). Prove \( \angle 1 \cong \angle 3 \).
    Show the full solution

    This is the second form of the congruent supplements theorem, supplementary to congruent angles rather than to the same angle. It can be cited directly, but proving it shows where the extra congruence enters. 1. \( \angle 1 \) and \( \angle 2 \) supplementary; \( \angle 3 \) and \( \angle 4 \) supplementary; \( \angle 2 \cong \angle 4 \). Reason: given. 2. \( m\angle 1 + m\angle 2 = 180 \). Reason: definition of supplementary angles. 3. \( m\angle 3 + m\angle 4 = 180 \). Reason: definition of supplementary angles. 4. \( m\angle 2 = m\angle 4 \). Reason: definition of congruent angles. 5. \( m\angle 3 + m\angle 2 = 180 \). Reason: substitution property of equality, using lines 3 and 4. 6. \( m\angle 1 + m\angle 2 = m\angle 3 + m\angle 2 \). Reason: transitive property of equality, using lines 2 and 5. 7. \( m\angle 1 = m\angle 3 \). Reason: subtraction property of equality. 8. \( \angle 1 \cong \angle 3 \). Reason: definition of congruent angles. The only new work is lines 4 and 5, which use the given congruence to make the two equations refer to the same angle. After that the proof is identical to the worked example. Proved in eight lines; the given congruence is used to make both equations share a term

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Name the three proof formats in this course.
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   Two-column, paragraph and flowchart

2. Does a paragraph proof need reasons?
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   Yes; every reason must appear in the prose

3. What do the arrows in a flowchart proof show?
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   Which statements follow from which

4. Which format best shows two independent branches converging?
   Show the full solution

   Flowchart

5. Is a correct paragraph proof worth full marks when the question does not specify a format?
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   Yes

6. Rewrite as a paragraph proof: given \( M \) is the midpoint of \( \overline{AB} \), prove \( AM = \frac{1}{2}AB \).
   Show the full solution

   Since \( M \) is the midpoint of \( \overline{AB} \), the definition of a midpoint gives \( \overline{AM} \cong \overline{MB} \), and therefore \( AM = MB \) by the definition of congruent segments. Because \( M \) lies between \( A \) and \( B \), the segment addition postulate gives \( AM + MB = AB \). Substituting \( AM \) for \( MB \) yields \( 2 \cdot AM = AB \), so \( AM = \frac{1}{2}AB \) by the division property of equality. Every reason from the seven-line two-column version survives: definition of midpoint, definition of congruent segments, segment addition postulate, substitution, division property. The paragraph is shorter only because the numbering is gone. Proved, with all five reasons named in the prose

7. A paragraph reads: "The angles are congruent because the triangles are congruent, and the triangles are congruent because the angles are congruent." Diagnose it.
   Show the full solution

   It is circular. Each claim is justified by the other, so the pair supports itself and rests on nothing established. Unwinding it: to accept the first sentence a reader must already accept that the triangles are congruent, and the only offered reason for that is the first sentence's conclusion. The argument never touches a given, a definition, a postulate or a previously proved theorem, which is what a proof is required to do. A correct version would establish triangle congruence from a criterion such as SAS, using given information, and then conclude the angles congruent by corresponding parts. The direction runs one way: from givens through a criterion to congruence and then to the parts. Circularity is easier to hide in a paragraph than in a two-column proof, which is exactly why the two-column format is taught first. In columns, both of these lines would need reasons, and neither reason would be available. Circular reasoning; each claim is the other's justification and neither rests on a given

8. Describe a flowchart proof for: given \( \angle 1 \cong \angle 2 \) and \( \angle 2 \cong \angle 3 \), prove \( \angle 1 \cong \angle 3 \).
   Show the full solution

   Two boxes at the top, side by side: "\( \angle 1 \cong \angle 2 \)" with the reason "given" beneath it, and "\( \angle 2 \cong \angle 3 \)" with "given" beneath it. An arrow from each leads into a single box below: "\( \angle 1 \cong \angle 3 \)" with the reason "transitive property of congruence" beneath it. The flowchart makes the structure obvious in a way the linear formats do not: both givens are needed simultaneously for the single inference, and neither leads anywhere on its own. A two-column proof of the same argument would list the givens as lines 1 and 2 and the conclusion as line 3, which implies a sequence that is not really there. Two given boxes converging by arrows on one conclusion box labeled with the transitive property

9. Write a paragraph proof: given \( \angle A \) and \( \angle B \) are both supplementary to \( \angle C \), prove \( \angle A \cong \angle B \).
   Show the full solution

   Since \( \angle A \) and \( \angle C \) are supplementary, the definition of supplementary angles gives \( m\angle A + m\angle C = 180 \). Since \( \angle B \) and \( \angle C \) are also supplementary, the same definition gives \( m\angle B + m\angle C = 180 \). Both expressions equal 180, so by the transitive property of equality \( m\angle A + m\angle C = m\angle B + m\angle C \). Subtracting \( m\angle C \) from both sides, by the subtraction property of equality, gives \( m\angle A = m\angle B \), and therefore \( \angle A \cong \angle B \) by the definition of congruent angles. This is the congruent supplements theorem, so in any later proof it could be cited in a single line by name rather than rebuilt. Writing it out here is what earns that right. Proved; this is the congruent supplements theorem

10. Explain when each format is the right choice, with a reason for each.
    Show the full solution

    Two-column. Best when the argument is a single chain of many steps, and best when learning, because the format makes an unjustified claim visible as an empty reason. Also the right choice whenever a question asks for a two-column proof, since format instructions are part of the question. The numbering makes it possible to refer back to a specific line, which matters in long proofs. Paragraph. Best for short arguments of three or four steps, where columns are more apparatus than the content needs. It is also how mathematicians actually write proofs, so it is worth becoming fluent in. The risk is that connectives can disguise a gap, so a paragraph proof must be checked by asking whether each claim still names its reason. Flowchart. Best when several independent facts are established separately and then combined, because the arrows show the structure directly. The triangle congruence proofs of unit 5 are the natural home for it: three separate pairs of congruent parts are each established on their own branch, and all three converge on a single application of SSS or SAS. A two-column version of the same proof lists nine lines in an order that suggests a sequence which is not really there. Chain of steps, two columns; short argument, paragraph; converging independent threads, flowchart

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use the numbering from the worked example.

1. What is a transversal?
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   A line crossing two or more lines at distinct points

2. How many angles does a transversal create when crossing two lines?
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   Eight

3. Which angles are interior in the worked example's numbering?
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   \( \angle 3 \), \( \angle 4 \), \( \angle 5 \), \( \angle 6 \)

4. Name the angle corresponding to \( \angle 2 \).
   Show the full solution

   \( \angle 6 \)

5. Are corresponding angles always congruent?
   Show the full solution

   Only when the lines are parallel

6. Classify the pair \( \angle 1 \) and \( \angle 7 \).
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   \( \angle 1 \) is exterior, at the upper intersection, on the left. \( \angle 7 \) is exterior, at the lower intersection, on the right. Both exterior, opposite sides of the transversal, different intersections. Alternate exterior angles

7. Classify the pair \( \angle 4 \) and \( \angle 5 \).
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   \( \angle 4 \) is interior, upper intersection, left side. \( \angle 5 \) is interior, lower intersection, left side. Both interior, same side of the transversal. Same-side interior angles

8. Two lines that are not parallel are cut by a transversal, and one pair of corresponding angles measures \( 70^\circ \) and \( 85^\circ \). Is the pair still called corresponding?
   Show the full solution

   Yes. The term describes where the angles sit relative to the two lines and the transversal, and that arrangement is unchanged by the measures. What has changed is that no theorem applies. The corresponding angles postulate has "if two parallel lines are cut by a transversal" as its hypothesis, and that hypothesis fails here, so the postulate says nothing. In fact the measures being unequal proves the lines are not parallel, by the contrapositive of the postulate. That is a legitimate inference and it runs in the opposite direction from the usual one. Yes; the name is positional, and the unequal measures instead prove the lines are not parallel

9. Explain why every theorem in the next lesson has parallelism in its hypothesis.
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   Because without it the conclusions are false. Draw two lines crossing a transversal at clearly different angles and every pair in the lesson has unequal measures. Parallelism is the condition that makes the two intersections identical in shape, and that identity is the source of every congruence in lesson 3.2. This is worth stating because the theorems are used so often that students stop reading their hypotheses. A diagram showing a transversal invites the reflex "so those angles are equal," and the reflex is wrong whenever the parallel marks are absent. The practical discipline: before citing any theorem from lesson 3.2, find where the parallelism came from. It is either given, marked on the diagram with arrowheads, or established earlier in the proof. If it is none of these, the theorem cannot be used and something else is needed. Without parallelism the conclusions are simply false, so the hypothesis is doing real work and must be located before the theorem is cited

10. Two lines are cut by a transversal, and \( \angle 3 \) and \( \angle 6 \) are same-side interior with \( m\angle 3 = 112^\circ \). What, if anything, follows about \( m\angle 6 \)?
    Show the full solution

    Nothing follows, because the problem never says the lines are parallel. If they were parallel, the same-side interior angles theorem of lesson 3.2 would give \( m\angle 6 = 180 - 112 = 68^\circ \). That is the answer most students write, and it is unjustified here. What can be said without parallelism: \( \angle 3 \) and \( \angle 4 \) form a linear pair at the upper intersection, so \( m\angle 4 = 68^\circ \) by the linear pair theorem. And \( \angle 3 \) is vertical to \( \angle 1 \), so \( m\angle 1 = 112^\circ \). Every angle at the upper intersection is determined, because those relationships hold at any intersection of two lines. Nothing at the lower intersection is determined, because nothing connects the two intersections. Parallelism is exactly what would supply that connection. Nothing about \( \angle 6 \); the four angles at \( \angle 3 \)'s own intersection are determined, but nothing links the two intersections without parallelism

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Assume \( \ell \parallel m \) throughout.

1. Corresponding angles measure \( 73^\circ \) and \( x^\circ \). Find \( x \).
   Show the full solution

   73

2. Alternate interior angles measure \( 105^\circ \) and \( y^\circ \). Find \( y \).
   Show the full solution

   105

3. Same-side interior angles measure \( 64^\circ \) and \( z^\circ \). Find \( z \).
   Show the full solution

   They are supplementary, not congruent. 116

4. Which of the four relationships gives supplementary rather than congruent angles?
   Show the full solution

   Same-side interior angles

5. How many distinct angle measures appear among the eight angles?
   Show the full solution

   Two, unless the transversal is perpendicular, in which case one

6. Two alternate exterior angles measure \( (3x + 14)^\circ \) and \( (5x - 26)^\circ \). Find both.
   Show the full solution

   Alternate exterior angles are congruent when the lines are parallel, so their measures are equal: \( 3x + 14 = 5x - 26 \), giving \( 40 = 2x \) and \( x = 20 \). Each measures \( 3(20) + 14 = 74^\circ \), confirmed by the other expression: \( 5(20) - 26 = 74^\circ \). Both \( 74^\circ \)

7. Two same-side interior angles measure \( (4x)^\circ \) and \( (2x + 30)^\circ \). Find both.
   Show the full solution

   Same-side interior angles are supplementary, so their measures sum to 180: \( 4x + (2x + 30) = 180 \), giving \( 6x = 150 \) and \( x = 25 \). The angles measure \( 4(25) = 100^\circ \) and \( 2(25) + 30 = 80^\circ \). Check: \( 100 + 80 = 180 \). Correct, and one is obtuse while the other is acute, as supplementary angles must be unless both are right. The trap here is setting the expressions equal, which would give \( x = 15 \) and two angles of \( 60^\circ \). Those are not supplementary, and the check catches it. \( 100^\circ \) and \( 80^\circ \)

8. Prove the alternate exterior angles theorem.
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   Given: \( \ell \parallel m \) cut by transversal \( t \). Prove: \( \angle 1 \cong \angle 7 \), an alternate exterior pair. 1. \( \ell \parallel m \). Reason: given. 2. \( \angle 1 \cong \angle 5 \). Reason: corresponding angles postulate. 3. \( \angle 5 \cong \angle 7 \). Reason: vertical angles theorem. 4. \( \angle 1 \cong \angle 7 \). Reason: transitive property of congruence. The structure is identical to the alternate interior proof: one move across using parallelism, one move around using vertical angles, then transitivity. Proved in four lines

9. A transversal is perpendicular to one of two parallel lines. Prove it is perpendicular to the other.
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   Given: \( \ell \parallel m \), transversal \( t \perp \ell \). Prove: \( t \perp m \). 1. \( \ell \parallel m \) and \( t \perp \ell \). Reason: given. 2. \( t \) and \( \ell \) form a right angle, call it \( \angle 1 \). Reason: definition of perpendicular lines. 3. \( m\angle 1 = 90 \). Reason: definition of a right angle. 4. \( \angle 1 \cong \angle 5 \), where \( \angle 5 \) is the corresponding angle at \( m \). Reason: corresponding angles postulate. 5. \( m\angle 5 = 90 \). Reason: definition of congruent angles and substitution. 6. \( \angle 5 \) is a right angle. Reason: definition of a right angle. 7. \( t \perp m \). Reason: definition of perpendicular lines. This is the perpendicular transversal theorem, which lesson 3.4 will cite by name. It is also a good illustration of a definition being used in both directions: line 2 uses it forward and line 7 uses it backward. Proved in seven lines

10. Three parallel lines are cut by a transversal. One angle at the top intersection measures \( 47^\circ \). Determine every angle in the figure and explain why only two values occur.
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    Three lines cut by a transversal give three intersections and twelve angles. At the top intersection. The given angle is \( 47^\circ \). Its linear pair partner is \( 180 - 47 = 133^\circ \). Its vertical angle is \( 47^\circ \), and the fourth is \( 133^\circ \). At the other intersections. Each of the four angles at the middle and bottom intersections corresponds to one at the top, since all three lines are parallel. By the corresponding angles postulate, each measures the same as its partner, so each intersection repeats the same pattern: two angles of \( 47^\circ \) and two of \( 133^\circ \). All twelve angles: six measure \( 47^\circ \) and six measure \( 133^\circ \). Why only two values. At any single intersection of two lines, only two values occur, because vertical angles are congruent and linear pairs are supplementary. Parallelism then forces every intersection to repeat the same pair, since corresponding angles are congruent. So the transversal meets every parallel line at the same pair of angles, however many lines there are. The check: \( 47 + 133 = 180 \), so the two values are supplementary, which is required for them to appear as a linear pair. Six angles of \( 47^\circ \) and six of \( 133^\circ \); parallelism makes every intersection identical

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Congruent corresponding angles prove what?
   Show the full solution

   The lines are parallel

2. What condition on same-side interior angles proves lines parallel?
   Show the full solution

   They are supplementary

3. Alternate interior angles measure \( 58^\circ \) and \( 58^\circ \). Are the lines parallel?
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   By the converse of the alternate interior angles theorem. Yes

4. Same-side interior angles measure \( 70^\circ \) and \( 100^\circ \). Are the lines parallel?
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   They sum to \( 170^\circ \), not \( 180^\circ \). No

5. What word must appear in the reason when proving lines parallel from angles?
   Show the full solution

   Converse

6. Alternate exterior angles measure \( (2x + 15)^\circ \) and \( (4x - 25)^\circ \). Find the \( x \) that makes the lines parallel.
   Show the full solution

   Alternate exterior angles are congruent exactly when the lines are parallel, by the theorem and its converse. So set them equal: \( 2x + 15 = 4x - 25 \), giving \( 40 = 2x \) and \( x = 20 \). Each angle then measures \( 2(20) + 15 = 55^\circ \), confirmed by \( 4(20) - 25 = 55^\circ \). \( x = 20 \), with both angles \( 55^\circ \)

7. Same-side interior angles measure \( (3x + 20)^\circ \) and \( (5x - 40)^\circ \). Find the \( x \) that makes the lines parallel.
   Show the full solution

   Same-side interior angles are supplementary exactly when the lines are parallel, so their measures must sum to 180: \( (3x + 20) + (5x - 40) = 180 \), giving \( 8x - 20 = 180 \), so \( 8x = 200 \) and \( x = 25 \). The angles measure \( 3(25) + 20 = 95^\circ \) and \( 5(25) - 40 = 85^\circ \). Check: \( 95 + 85 = 180 \). Correct. The trap is setting them equal, which gives \( x = 30 \) and two angles of \( 110^\circ \). Those sum to \( 220^\circ \), so the lines would not be parallel. The check catches it. \( x = 25 \)

8. Explain why using the alternate interior angles theorem to prove lines parallel is circular.
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   The theorem states: if two parallel lines are cut by a transversal, then alternate interior angles are congruent. Parallelism is its hypothesis. A proof aiming to establish parallelism cannot cite a theorem that assumes parallelism, because the hypothesis is precisely what is not yet available. Doing so assumes the conclusion, which is the second of the four errors this course names. What is needed instead is the converse, proved separately: if alternate interior angles are congruent, then the lines are parallel. Its hypothesis is the congruence, which the problem supplies, and its conclusion is the parallelism, which is wanted. Both statements are true here, which is exactly what makes the error easy to commit and hard to notice. A student who thinks of them as one fact will write the wrong reason and reach the right answer, and will have no defense when a theorem's converse turns out to be false. The theorem takes parallelism as its hypothesis, so citing it assumes what is being proved; the converse is the theorem needed

9. \( \angle 1 \cong \angle 7 \), alternate exterior angles. Prove \( \ell \parallel m \) two different ways.
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   Route one, directly. 1. \( \angle 1 \cong \angle 7 \). Reason: given. 2. \( \ell \parallel m \). Reason: converse of the alternate exterior angles theorem. Route two, through corresponding angles. 1. \( \angle 1 \cong \angle 7 \). Reason: given. 2. \( \angle 7 \cong \angle 5 \). Reason: vertical angles theorem. 3. \( \angle 1 \cong \angle 5 \). Reason: transitive property of congruence. 4. \( \ell \parallel m \). Reason: converse of the corresponding angles postulate, since \( \angle 1 \) and \( \angle 5 \) are corresponding. Both are valid. The first is shorter because it uses a theorem tailored to the given pair; the second shows that all four converses are really one fact reached by different routes. Either earns full marks. Proved, directly in two lines or through corresponding angles in four

10. In a figure, \( \ell \parallel m \) and \( m \parallel n \). Prove \( \ell \parallel n \), and explain why this needs a proof at all.
    Show the full solution

    Given: \( \ell \parallel m \) and \( m \parallel n \), all cut by a transversal \( t \). Prove: \( \ell \parallel n \). 1. \( \ell \parallel m \) and \( m \parallel n \). Reason: given. 2. \( \angle 1 \cong \angle 2 \), where \( \angle 1 \) is at \( \ell \) and \( \angle 2 \) is the corresponding angle at \( m \). Reason: corresponding angles postulate. 3. \( \angle 2 \cong \angle 3 \), where \( \angle 3 \) is the corresponding angle at \( n \). Reason: corresponding angles postulate. 4. \( \angle 1 \cong \angle 3 \). Reason: transitive property of congruence. 5. \( \ell \parallel n \). Reason: converse of the corresponding angles postulate. Why it needs proving. The result is obvious enough that it is easy to assume, but "parallel" is defined as never meeting, and nothing in that definition immediately gives transitivity. Two lines each failing to meet a third could in principle meet each other; that is what happens with the analogous statement for perpendicularity, where two lines perpendicular to the same line are parallel rather than perpendicular to each other. The proof also illustrates the pattern of this unit: the forward direction is used twice to extract angle facts, and the converse is used once at the end to convert those facts back into parallelism. Both directions appear in one proof, and the reason column is what keeps them apart. Proved in five lines; transitivity of parallelism does not follow from the definition alone and the analogous claim for perpendicularity is false

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. If \( t \perp \ell \) and \( \ell \parallel m \), what follows about \( t \) and \( m \)?
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   \( t \perp m \), by the perpendicular transversal theorem

2. If \( a \perp c \) and \( b \perp c \), what follows about \( a \) and \( b \)?
   Show the full solution

   \( a \parallel b \)

3. Is perpendicularity transitive?
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   No

4. Is parallelism transitive?
   Show the full solution

   Yes

5. Two perpendicular lines meet. What is the measure of each of the four angles?
   Show the full solution

   \( 90^\circ \)

6. Given \( \ell \parallel m \) and \( t \perp \ell \), find every angle in the figure.
   Show the full solution

   Since \( t \perp \ell \), all four angles at the \( \ell \) intersection are right angles, measuring \( 90^\circ \) each, by the definition of perpendicular lines and the fact that one right angle forces all four. By the perpendicular transversal theorem, \( t \perp m \) as well, so all four angles at the \( m \) intersection measure \( 90^\circ \). All eight angles measure \( 90^\circ \). This is the exceptional case noted in lesson 3.2: normally two distinct values appear among the eight angles, but when the transversal is perpendicular, the two values coincide because \( 180 - 90 = 90 \). All eight measure \( 90^\circ \)

7. In a figure, \( p \perp q \) and \( q \perp r \) and \( r \perp s \). Find the relationship between \( p \) and \( s \).
   Show the full solution

   Work along the chain one step at a time. From \( p \perp q \) and \( q \perp r \): the two lines perpendicular to \( q \) are parallel, so \( p \parallel r \). From \( p \parallel r \) and \( r \perp s \): the perpendicular transversal theorem gives \( p \perp s \), since \( s \) is perpendicular to one of two parallel lines. Coordinate check: let \( q \) be \( y = 0 \), so \( p \) is vertical, say \( x = 0 \). Then \( r \perp q \) makes \( r \) vertical, say \( x = 4 \). Then \( s \perp r \) makes \( s \) horizontal, say \( y = 7 \). And \( x = 0 \) is perpendicular to \( y = 7 \). Confirmed. The pattern is worth noticing: each perpendicular flips between two directions, so an even number of them returns to the start and an odd number does not. Three perpendiculars give perpendicular; two give parallel. \( p \perp s \)

8. Explain why establishing one right angle is enough to prove two lines perpendicular.
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   Two intersecting lines form four angles, and they are not independent. Suppose one of them, \( \angle 1 \), is a right angle, so \( m\angle 1 = 90 \). The angle adjacent to it forms a linear pair with it, so the two are supplementary by the linear pair theorem, giving the adjacent angle a measure of \( 180 - 90 = 90 \). The same argument applies to the other adjacent angle. The fourth angle is vertical to \( \angle 1 \), so it is congruent to it by the vertical angles theorem and also measures \( 90 \). So all four are right angles as soon as one is. The definition of perpendicular only asks that the lines meet at a right angle, and a proof therefore needs to establish exactly one. This is a small efficiency with a large effect on proof length, since the alternative would be four separate arguments every time perpendicularity is claimed. A linear pair and a vertical angle force the other three to be right as soon as one is

9. Prove the perpendicular transversal theorem using alternate interior angles rather than corresponding angles.
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   Given: \( \ell \parallel m \), \( t \perp \ell \). Prove: \( t \perp m \). 1. \( \ell \parallel m \) and \( t \perp \ell \). Reason: given. 2. \( \angle 3 \), an interior angle at \( \ell \), is a right angle. Reason: definition of perpendicular lines, using the fact that all four angles at that intersection are right. 3. \( \angle 3 \cong \angle 5 \), the alternate interior angle at \( m \). Reason: alternate interior angles theorem. 4. \( m\angle 3 = m\angle 5 \). Reason: definition of congruent angles. 5. \( m\angle 3 = 90 \). Reason: definition of a right angle. 6. \( m\angle 5 = 90 \). Reason: substitution property of equality. 7. \( \angle 5 \) is a right angle. Reason: definition of a right angle. 8. \( t \perp m \). Reason: definition of perpendicular lines. The route through alternate interior angles is one line longer than the corresponding angles route, because the interior angle at \( \ell \) has to be identified as right first. Both are valid; when a question does not specify, choose the shorter. Proved in eight lines

10. Lines \( a \) and \( b \) are both perpendicular to \( c \), and \( a \) passes through \( (0, 0) \) while \( b \) passes through \( (0, 5) \). Line \( c \) has slope 2. Find the equations of \( a \) and \( b \), and confirm they are parallel.
    Show the full solution

    Since \( a \perp c \) and \( c \) has slope 2, the slope of \( a \) is the negative reciprocal, \( -\dfrac{1}{2} \). The same reasoning gives \( b \) the same slope. \( a \) passes through the origin, so \( a: y = -\dfrac{1}{2}x \). \( b \) passes through \( (0, 5) \), so \( b: y = -\dfrac{1}{2}x + 5 \). They have equal slopes and different intercepts, so they are distinct parallel lines, which is exactly what the two-perpendiculars theorem predicted. Check the perpendicularity: \( 2 \times \left(-\dfrac{1}{2}\right) = -1 \), so each is genuinely perpendicular to \( c \). This is the coordinate version of the theorem, and it shows why the theorem is true in a second way: perpendicularity to a line of slope \( m \) forces the slope \( -\dfrac{1}{m} \), and any two lines forced to the same slope are parallel. Lesson 3.5 develops that connection properly. \( a: y = -\frac{1}{2}x \) and \( b: y = -\frac{1}{2}x + 5 \), parallel because both slopes are \( -\frac{1}{2} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What slope is parallel to a line of slope \( -4 \)?
   Show the full solution

   \( -4 \)

2. What slope is perpendicular to a line of slope \( -4 \)?
   Show the full solution

   The negative reciprocal. \( \frac{1}{4} \)

3. Are \( y = 3x + 1 \) and \( y = 3x - 7 \) parallel?
   Show the full solution

   Equal slopes, different intercepts. Yes

4. Are \( y = \frac{2}{3}x \) and \( y = -\frac{3}{2}x + 4 \) perpendicular?
   Show the full solution

   \( \frac{2}{3} \times \left(-\frac{3}{2}\right) = -1 \). Yes

5. What is the slope of a vertical line?
   Show the full solution

   Undefined

6. Classify \( 3x + y = 7 \) and \( x - 3y = 9 \).
   Show the full solution

   First: \( y = -3x + 7 \), so the slope is \( -3 \). Second: \( -3y = -x + 9 \), so \( y = \dfrac{1}{3}x - 3 \) and the slope is \( \dfrac{1}{3} \). Product: \( -3 \times \dfrac{1}{3} = -1 \). Perpendicular

7. Classify \( 2x - 5y = 10 \) and \( 4x - 10y = 3 \).
   Show the full solution

   First: \( -5y = -2x + 10 \), so \( y = \dfrac{2}{5}x - 2 \), slope \( \dfrac{2}{5} \). Second: \( -10y = -4x + 3 \), so \( y = \dfrac{2}{5}x - \dfrac{3}{10} \), slope \( \dfrac{2}{5} \). Equal slopes. Different intercepts, \( -2 \) against \( -0.3 \), so they are distinct lines rather than the same one. Parallel

8. A quadrilateral has vertices \( A(0,0) \), \( B(4,2) \), \( C(6,6) \), \( D(2,4) \). Show that \( \overline{AB} \parallel \overline{DC} \) and \( \overline{AD} \parallel \overline{BC} \).
   Show the full solution

   Slope of \( \overline{AB} \): \( \dfrac{2 - 0}{4 - 0} = \dfrac{1}{2} \). Slope of \( \overline{DC} \): \( \dfrac{6 - 4}{6 - 2} = \dfrac{2}{4} = \dfrac{1}{2} \). Equal, so \( \overline{AB} \parallel \overline{DC} \). Slope of \( \overline{AD} \): \( \dfrac{4 - 0}{2 - 0} = 2 \). Slope of \( \overline{BC} \): \( \dfrac{6 - 2}{6 - 4} = \dfrac{4}{2} = 2 \). Equal, so \( \overline{AD} \parallel \overline{BC} \). Both pairs of opposite sides are parallel, so \( ABCD \) is a parallelogram by definition. This is the coordinate method that unit 7 develops fully. Note also that \( \frac{1}{2} \times 2 = 1 \), not \( -1 \), so adjacent sides are not perpendicular and the figure is not a rectangle. Both pairs of opposite sides have equal slopes, so the figure is a parallelogram

9. Explain why the product of perpendicular slopes is \( -1 \), and why a vertical and horizontal pair is an exception.
   Show the full solution

   Take a line of slope \( m = \dfrac{a}{b} \), meaning it rises \( a \) for a run of \( b \). Draw the right triangle with legs \( b \) horizontal and \( a \) vertical along this line. Rotating that triangle a quarter turn about the point where the lines meet produces a triangle whose horizontal leg is \( a \) and whose vertical leg is \( b \), with the direction reversed. The rotated line has slope \( -\dfrac{b}{a} \). Rotating by a quarter turn is exactly what makes the lines perpendicular. So the perpendicular slope is \( -\dfrac{b}{a} \), and the product is \( \dfrac{a}{b} \times \left(-\dfrac{b}{a}\right) = -1 \). The exception. The argument assumed both \( a \) and \( b \) are nonzero. A horizontal line has \( a = 0 \), giving slope 0, and its perpendicular would need slope \( -\dfrac{b}{0} \), which is undefined, and indeed a vertical line has undefined slope. The product \( 0 \times \text{undefined} \) is not a number, so the test cannot be applied even though the lines are genuinely perpendicular. This is why the rule is always stated for non-vertical, non-horizontal lines, and why a classification problem should check for a vertical line before reaching for the product test. A quarter-turn rotation swaps rise and run and reverses one sign; the rule fails when a slope is 0 or undefined because the product is not a number

10. Triangle \( PQR \) has vertices \( P(1,1) \), \( Q(5,3) \), \( R(3,7) \). Determine whether it has a right angle.
    Show the full solution

    A right angle at a vertex means the two sides meeting there are perpendicular, so compute all three slopes and test each pair. Slope of \( \overline{PQ} \): \( \dfrac{3-1}{5-1} = \dfrac{2}{4} = \dfrac{1}{2} \). Slope of \( \overline{QR} \): \( \dfrac{7-3}{3-5} = \dfrac{4}{-2} = -2 \). Slope of \( \overline{PR} \): \( \dfrac{7-1}{3-1} = \dfrac{6}{2} = 3 \). At \( Q \), the sides are \( \overline{PQ} \) and \( \overline{QR} \): \( \dfrac{1}{2} \times (-2) = -1 \). Perpendicular, so \( \angle Q \) is a right angle. Checking the others for completeness. At \( P \): \( \dfrac{1}{2} \times 3 = \dfrac{3}{2} \), not \( -1 \). At \( R \): \( -2 \times 3 = -6 \), not \( -1 \). So exactly one right angle, at \( Q \), and the triangle is right. A second check by lengths: \( PQ = \sqrt{16 + 4} = \sqrt{20} \), \( QR = \sqrt{4 + 16} = \sqrt{20} \), \( PR = \sqrt{4 + 36} = \sqrt{40} \). Then \( PQ^2 + QR^2 = 20 + 20 = 40 = PR^2 \), which is the converse of the Pythagorean theorem confirming a right angle at \( Q \). The two legs are also equal, so the triangle is right isosceles. Yes, a right angle at \( Q \), confirmed by both the slope test and the Pythagorean converse

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Write the line through \( (0, 4) \) parallel to \( y = 5x \).
   Show the full solution

   \( y = 5x + 4 \)

2. Write the line through \( (1, 2) \) parallel to \( y = -x + 9 \).
   Show the full solution

   \( y - 2 = -(x - 1) \). \( y = -x + 3 \)

3. Write the line through \( (0, 0) \) perpendicular to \( y = 4x \).
   Show the full solution

   Negative reciprocal of 4. \( y = -\frac{1}{4}x \)

4. Write the line through \( (5, 3) \) perpendicular to \( x = 5 \).
   Show the full solution

   The reference is vertical, so the answer is horizontal through \( y = 3 \). \( y = 3 \)

5. What slope is perpendicular to \( -\frac{5}{2} \)?
   Show the full solution

   \( \frac{2}{5} \)

6. Write the line through \( (-2, 7) \) parallel to \( 3x - y = 8 \).
   Show the full solution

   Rearrange: \( -y = -3x + 8 \), so \( y = 3x - 8 \) and the slope is 3. Parallel means slope 3. Point-slope with \( (-2, 7) \): \( y - 7 = 3(x + 2) \), so \( y - 7 = 3x + 6 \) and \( y = 3x + 13 \). Check: at \( x = -2 \), \( y = -6 + 13 = 7 \). Correct. \( y = 3x + 13 \)

7. Write the line through \( (6, -4) \) perpendicular to \( 4x - 2y = 9 \).
   Show the full solution

   Rearrange: \( -2y = -4x + 9 \), so \( y = 2x - \dfrac{9}{2} \) and the slope is 2. Perpendicular slope: \( -\dfrac{1}{2} \). Check: \( 2 \times \left(-\dfrac{1}{2}\right) = -1 \). Point-slope with \( (6, -4) \): \( y + 4 = -\dfrac{1}{2}(x - 6) \), so \( y + 4 = -\dfrac{1}{2}x + 3 \) and \( y = -\dfrac{1}{2}x - 1 \). Check: at \( x = 6 \), \( y = -3 - 1 = -4 \). Correct. \( y = -\frac{1}{2}x - 1 \)

8. Find the perpendicular bisector of the segment from \( (1, 3) \) to \( (9, 7) \).
   Show the full solution

   Two things are needed: the midpoint and the perpendicular slope. Midpoint: \( \left( \dfrac{1+9}{2}, \dfrac{3+7}{2} \right) = (5, 5) \). Slope of the segment: \( \dfrac{7-3}{9-1} = \dfrac{4}{8} = \dfrac{1}{2} \). Perpendicular slope: \( -2 \). Check: \( \dfrac{1}{2} \times (-2) = -1 \). Equation: \( y - 5 = -2(x - 5) \), so \( y = -2x + 15 \). Check both conditions: at \( x = 5 \), \( y = -10 + 15 = 5 \), so it passes through the midpoint; and its slope is the negative reciprocal of the segment's. Both halves of the definition hold. \( y = -2x + 15 \)

9. A line passes through \( (2, 1) \) and is perpendicular to the line through \( (0, 4) \) and \( (6, 0) \). Find its equation.
   Show the full solution

   First find the reference slope from its two points: \( m = \dfrac{0 - 4}{6 - 0} = \dfrac{-4}{6} = -\dfrac{2}{3} \). Perpendicular slope: flip and change sign, giving \( \dfrac{3}{2} \). Check: \( -\dfrac{2}{3} \times \dfrac{3}{2} = -1 \). Point-slope with \( (2, 1) \): \( y - 1 = \dfrac{3}{2}(x - 2) \), so \( y - 1 = \dfrac{3}{2}x - 3 \) and \( y = \dfrac{3}{2}x - 2 \). Check: at \( x = 2 \), \( y = 3 - 2 = 1 \). Correct. The step worth noting is that the reference line was given by two points rather than an equation, so the slope had to be computed before anything else. The rest of the procedure is unchanged. \( y = \frac{3}{2}x - 2 \)

10. A rectangle has three vertices at \( A(0,0) \), \( B(6,2) \) and \( C(4,8) \)? Determine whether this is possible, and if so find the fourth vertex.
    Show the full solution

    A rectangle needs right angles, so first check whether \( \angle B \) is right, since \( B \) is between \( A \) and \( C \) in the listing. Slope of \( \overline{AB} \): \( \dfrac{2-0}{6-0} = \dfrac{1}{3} \). Slope of \( \overline{BC} \): \( \dfrac{8-2}{4-6} = \dfrac{6}{-2} = -3 \). Product: \( \dfrac{1}{3} \times (-3) = -1 \). Perpendicular, so \( \angle B \) is a right angle and a rectangle is possible. The fourth vertex \( D \) must make \( \overline{AD} \parallel \overline{BC} \) and \( \overline{DC} \parallel \overline{AB} \). Since \( ABCD \) is a parallelogram, the diagonals bisect each other, so the midpoint of \( \overline{AC} \) equals the midpoint of \( \overline{BD} \). Midpoint of \( \overline{AC} \): \( \left( \dfrac{0+4}{2}, \dfrac{0+8}{2} \right) = (2, 4) \). So \( \left( \dfrac{6 + x}{2}, \dfrac{2 + y}{2} \right) = (2, 4) \), giving \( 6 + x = 4 \) and \( 2 + y = 8 \), so \( x = -2 \) and \( y = 6 \). \( D \) is \( (-2, 6) \). Check: slope of \( \overline{AD} \) is \( \dfrac{6-0}{-2-0} = -3 \), matching \( \overline{BC} \). Slope of \( \overline{DC} \) is \( \dfrac{8-6}{4-(-2)} = \dfrac{2}{6} = \dfrac{1}{3} \), matching \( \overline{AB} \). Both pairs of opposite sides are parallel and adjacent sides are perpendicular, so \( ABCD \) is a rectangle. Possible; the fourth vertex is \( D(-2, 6) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What path gives the distance from a point to a line?
   Show the full solution

   The perpendicular segment

2. Find the distance from \( (3, 7) \) to the line \( y = 2 \).
   Show the full solution

   The line is horizontal, so the perpendicular is vertical. 5

3. Find the distance from \( (6, 1) \) to the line \( x = -2 \).
   Show the full solution

   8

4. Find the distance between \( y = 4 \) and \( y = 11 \).
   Show the full solution

   7

5. Is the vertical gap from a point to a line the same as the distance?
   Show the full solution

   Only if the line is horizontal; otherwise it is larger

6. Find the distance from \( (0, 0) \) to the line \( 3x + 4y = 25 \).
   Show the full solution

   Using the formula with the line written \( 3x + 4y - 25 = 0 \): \( d = \dfrac{|3(0) + 4(0) - 25|}{\sqrt{9 + 16}} = \dfrac{25}{5} = 5 \). Verifying by construction: the line has slope \( -\dfrac{3}{4} \), so the perpendicular through the origin has slope \( \dfrac{4}{3} \), giving \( y = \dfrac{4}{3}x \). Substituting into \( 3x + 4y = 25 \): \( 3x + \dfrac{16x}{3} = 25 \), so \( 9x + 16x = 75 \) and \( x = 3 \), giving \( y = 4 \). Distance from \( (0,0) \) to \( (3,4) \) is \( \sqrt{9 + 16} = 5 \). The two methods agree. 5

7. Find the distance from \( (1, 2) \) to the line \( y = x \).
   Show the full solution

   Write the line as \( x - y = 0 \) and use the formula: \( d = \dfrac{|1 - 2|}{\sqrt{1 + 1}} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} \approx 0.707 \). By construction: the perpendicular through \( (1,2) \) has slope \( -1 \), so \( y - 2 = -(x - 1) \), giving \( y = -x + 3 \). Intersecting with \( y = x \): \( x = -x + 3 \), so \( x = 1.5 \) and \( y = 1.5 \). Distance from \( (1,2) \) to \( (1.5, 1.5) \): differences 0.5 and \( -0.5 \), so \( \sqrt{0.25 + 0.25} = \sqrt{0.5} = \dfrac{\sqrt{2}}{2} \). The two agree. \( \frac{\sqrt{2}}{2} \), about 0.71

8. Find the distance between \( y = 3x - 2 \) and \( y = 3x + 8 \).
   Show the full solution

   The lines are parallel, both with slope 3, so the distance is well defined. Take the point \( (0, -2) \) on the first line. Write the second as \( 3x - y + 8 = 0 \) and apply the formula: \( d = \dfrac{|3(0) - (-2) + 8|}{\sqrt{9 + 1}} = \dfrac{10}{\sqrt{10}} = \sqrt{10} \approx 3.16 \). Note that the vertical gap between the lines is \( 8 - (-2) = 10 \), while the actual distance is about 3.16. The steeper the lines, the larger the discrepancy, since the perpendicular is far from vertical. \( \sqrt{10} \), about 3.16

9. Explain why the perpendicular segment is the shortest path from a point to a line.
   Show the full solution

   Let \( P \) be the point, \( \ell \) the line, and \( F \) the foot of the perpendicular from \( P \) to \( \ell \). Let \( Q \) be any other point on \( \ell \). Since \( \overline{PF} \perp \ell \), the angle \( \angle PFQ \) is a right angle, so \( \triangle PFQ \) is a right triangle with legs \( \overline{PF} \) and \( \overline{FQ} \), and hypotenuse \( \overline{PQ} \). By the Pythagorean theorem, \( PQ^2 = PF^2 + FQ^2 \). Since \( Q \neq F \), the length \( FQ \) is positive, so \( FQ^2 \gt 0 \) and therefore \( PQ^2 \gt PF^2 \), giving \( PQ \gt PF \). So every other segment from \( P \) to the line is strictly longer than the perpendicular one. The perpendicular is not merely one candidate among several; it is the unique shortest, which is why "the distance" is well defined. The same argument proves the analogous fact in unit 6: in any right triangle the hypotenuse is the longest side, since it is opposite the largest angle. Any other segment is the hypotenuse of a right triangle with the perpendicular as a leg, and a hypotenuse always exceeds a leg

10. Find the distance from \( (5, 5) \) to the line through \( (1, 1) \) and \( (7, 4) \), by construction and by formula.
    Show the full solution

    Find the line first. Slope: \( \dfrac{4 - 1}{7 - 1} = \dfrac{3}{6} = \dfrac{1}{2} \). Through \( (1,1) \): \( y - 1 = \dfrac{1}{2}(x - 1) \), so \( y = \dfrac{1}{2}x + \dfrac{1}{2} \), or in standard form \( x - 2y + 1 = 0 \). By formula. \( d = \dfrac{|5 - 2(5) + 1|}{\sqrt{1 + 4}} = \dfrac{|5 - 10 + 1|}{\sqrt{5}} = \dfrac{4}{\sqrt{5}} = \dfrac{4\sqrt{5}}{5} \approx 1.789 \). By construction. The perpendicular through \( (5,5) \) has slope \( -2 \): \( y - 5 = -2(x - 5) \), so \( y = -2x + 15 \). Intersecting: \( \dfrac{1}{2}x + \dfrac{1}{2} = -2x + 15 \). Multiplying by 2: \( x + 1 = -4x + 30 \), so \( 5x = 29 \) and \( x = 5.8 \), giving \( y = -11.6 + 15 = 3.4 \). Distance from \( (5,5) \) to \( (5.8, 3.4) \): differences 0.8 and \( -1.6 \), so \( \sqrt{0.64 + 2.56} = \sqrt{3.2} \approx 1.789 \). The two methods agree. Note again that the foot of the perpendicular has untidy coordinates while the exact answer, \( \dfrac{4\sqrt{5}}{5} \), is compact. That is typical and is not a sign of arithmetic error. \( \frac{4\sqrt{5}}{5} \), about 1.79

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does a rigid motion preserve?
   Show the full solution

   Distance, and therefore angle measure

2. Name the three rigid motions.
   Show the full solution

   Translation, reflection, rotation

3. Is a dilation a rigid motion?
   Show the full solution

   No; it changes distances

4. What is the image of \( A \) usually called?
   Show the full solution

   \( A' \), read "A prime"

5. Which rigid motion reverses orientation?
   Show the full solution

   Reflection

6. Is \( (x, y) \to (x + 5, y - 2) \) a rigid motion? Test it on two points.
   Show the full solution

   Take \( P(0,0) \) and \( Q(3,4) \). Their distance is \( \sqrt{9 + 16} = 5 \). Images: \( P'(5, -2) \) and \( Q'(8, 2) \). Their distance is \( \sqrt{(8-5)^2 + (2-(-2))^2} = \sqrt{9 + 16} = 5 \). Unchanged. In general the rule adds the same amounts to both points, so the differences \( x_2 - x_1 \) and \( y_2 - y_1 \) are untouched and the distance formula gives the same result. Yes, a rigid motion; it is a translation

7. Is \( (x, y) \to (3x, y) \) a rigid motion?
   Show the full solution

   Take \( P(0,0) \) and \( Q(1,0) \), distance 1. Images: \( P'(0,0) \) and \( Q'(3,0) \), distance 3. Changed, so it is not rigid. This one is worth examining further. Take \( R(0,1) \): its image is \( R'(0,1) \), unchanged, so vertical distances survive while horizontal ones triple. The transformation stretches the plane in one direction only, which distorts shape as well as size: a square becomes a rectangle and the right angles survive but a \( 45^\circ \) angle does not. So it is neither rigid nor a dilation. A dilation multiplies all distances by the same factor; this multiplies only some. No; it stretches horizontally only, so it preserves neither distance nor shape

8. Explain why preserving distance forces preserving angle measure.
   Show the full solution

   Take an angle \( \angle ABC \) and mark points \( A \) and \( C \) on its two sides. The three points \( A \), \( B \), \( C \) form a triangle, and the angle at \( B \) is the one in question. A rigid motion sends these to \( A' \), \( B' \), \( C' \) with \( A'B' = AB \), \( B'C' = BC \) and \( A'C' = AC \). So the image triangle has three sides congruent to the original's. By the SSS criterion of lesson 5.2, the two triangles are congruent, and corresponding angles of congruent triangles are congruent. In particular \( \angle A'B'C' \cong \angle ABC \). So the angle measure survives, not as an extra assumption but as a consequence of the distances surviving. This is why the definition of a rigid motion mentions distance only: everything else follows. Three preserved distances make the triangles congruent by SSS, so corresponding angles are congruent

9. A transformation sends \( (1,2) \) to \( (1,-2) \) and \( (4,3) \) to \( (4,-3) \). Identify it and verify it is rigid.
   Show the full solution

   The pattern is that \( x \) stays the same and \( y \) changes sign: \( (x, y) \to (x, -y) \). That is reflection across the \( x \) axis, covered in lesson 4.3. Verifying rigidity. Original distance from \( (1,2) \) to \( (4,3) \): differences 3 and 1, so \( \sqrt{9 + 1} = \sqrt{10} \). Image distance from \( (1,-2) \) to \( (4,-3) \): differences 3 and \( -1 \), so \( \sqrt{9 + 1} = \sqrt{10} \). Unchanged. In general, the horizontal difference is untouched and the vertical difference has its sign flipped, which the squaring in the distance formula removes. So every distance is preserved and the transformation is rigid. Orientation, however, is reversed: a point above the axis moves below it, so a figure traced counterclockwise becomes one traced clockwise. That is the signature of a reflection. Reflection across the \( x \) axis; rigid, since the sign change is removed by squaring

10. A square has vertices \( (0,0) \), \( (2,0) \), \( (2,2) \), \( (0,2) \). Apply \( (x,y) \to (x + y, y) \) and determine what is and is not preserved.
    Show the full solution

    Apply the rule to each vertex. \( (0,0) \to (0,0) \); \( (2,0) \to (2,0) \); \( (2,2) \to (4,2) \); \( (0,2) \to (2,2) \). Distance. The bottom side from \( (0,0) \) to \( (2,0) \) had length 2 and still does. The left side from \( (0,0) \) to \( (0,2) \) had length 2; its image runs from \( (0,0) \) to \( (2,2) \), with length \( \sqrt{4 + 4} = 2\sqrt{2} \approx 2.83 \). Changed, so the transformation is not rigid. Angle measure. The original angle at \( (0,0) \) is a right angle between a horizontal and a vertical side. In the image the sides run along \( (2,0) \) and \( (2,2) \) directions, meeting at \( 45^\circ \). Not preserved. What is preserved. Parallel sides stay parallel: the bottom and top both remain horizontal, and the two slanted sides both have slope 1. Area is also preserved, since the base 2 and the height 2 are unchanged, giving 4 in both cases. So the image is a parallelogram of the same area but a different shape. This transformation is called a shear, and it shows that the categories in this lesson are not exhaustive: there are transformations that are neither rigid nor dilations, and preserving area does not imply preserving distance. A shear: parallelism and area are preserved, distance and angle measure are not, so the square becomes a non-rectangular parallelogram

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Apply \( (x,y) \to (x+3, y+2) \) to \( (1, 4) \).
   Show the full solution

   \( (4, 6) \)

2. Apply the vector \( \langle -2, 5 \rangle \) to \( (7, 0) \).
   Show the full solution

   \( (5, 5) \)

3. \( A(2, 3) \) maps to \( A'(6, 3) \). Find the vector.
   Show the full solution

   Image minus preimage. \( \langle 4, 0 \rangle \)

4. Does a translation change orientation?
   Show the full solution

   No

5. How many fixed points does a translation by \( \langle 3, 1 \rangle \) have?
   Show the full solution

   Every point moves. None

6. Translate the triangle with vertices \( (0,0) \), \( (4,0) \), \( (0,3) \) by \( \langle -1, 6 \rangle \), and confirm the side lengths are unchanged.
   Show the full solution

   Images: \( (-1, 6) \), \( (3, 6) \), \( (-1, 9) \). Original sides: from \( (0,0) \) to \( (4,0) \) is 4; from \( (0,0) \) to \( (0,3) \) is 3; from \( (4,0) \) to \( (0,3) \) is \( \sqrt{16 + 9} = 5 \). Image sides: from \( (-1,6) \) to \( (3,6) \) is 4; from \( (-1,6) \) to \( (-1,9) \) is 3; from \( (3,6) \) to \( (-1,9) \) is \( \sqrt{16 + 9} = 5 \). All three unchanged, confirming the translation is rigid. Images \( (-1,6) \), \( (3,6) \), \( (-1,9) \); sides 3, 4, 5 in both

7. A translation maps \( (5, -2) \) to \( (1, 3) \). Find the image of \( (0, 0) \).
   Show the full solution

   Find the vector first: \( \langle 1 - 5, \; 3 - (-2) \rangle = \langle -4, 5 \rangle \). Apply it to the origin: \( (0 - 4, 0 + 5) = (-4, 5) \). Note the shortcut: the image of the origin under a translation is always the vector itself, since the rule adds the vector's components to zero. \( (-4, 5) \)

8. What single translation undoes \( \langle 7, -4 \rangle \)?
   Show the full solution

   To return every point to where it started, move back by the same amount in the opposite direction, which negates both components: \( \langle -7, 4 \rangle \). Check by composing: a point \( (x,y) \) goes to \( (x + 7, y - 4) \) and then to \( (x + 7 - 7, y - 4 + 4) = (x, y) \), the identity. The vector sum is \( \langle 0, 0 \rangle \), which is the translation that moves nothing. Every rigid motion has an inverse, and for a translation it is simply the opposite vector. \( \langle -7, 4 \rangle \)

9. Explain why every segment joining a point to its translated image is parallel to every other such segment.
   Show the full solution

   Take a translation by \( \langle a, b \rangle \) and any point \( P(x, y) \). Its image is \( P'(x + a, y + b) \), so the segment \( \overline{PP'} \) has horizontal change \( a \) and vertical change \( b \). Those numbers do not depend on \( P \). Every point in the plane produces a segment with the same horizontal and vertical changes, so every such segment has slope \( \dfrac{b}{a} \), the same for all of them. Equal slopes mean parallel lines, so all the connecting segments are parallel. They also all have length \( \sqrt{a^2 + b^2} \), again independent of \( P \), so they are congruent as well. That is the geometric definition of a translation: a transformation for which all the connecting segments are parallel and congruent. The coordinate rule and the geometric description are two views of the same thing. The horizontal and vertical changes are the same for every point, so all the segments have the same slope and length

10. A translation maps \( \triangle ABC \) to \( \triangle A'B'C' \), where \( A(1,2) \), \( B(5,2) \), \( C(1,8) \) and \( A'(4, -1) \). Find \( B' \), \( C' \), and the perimeter of each triangle.
    Show the full solution

    Find the vector from the one pair given: \( \langle 4 - 1, \; -1 - 2 \rangle = \langle 3, -3 \rangle \). Apply it to the other vertices. \( B(5,2) \to B'(8, -1) \). \( C(1,8) \to C'(4, 5) \). Original perimeter. \( AB \): from \( (1,2) \) to \( (5,2) \), length 4. \( AC \): from \( (1,2) \) to \( (1,8) \), length 6. \( BC \): from \( (5,2) \) to \( (1,8) \), differences \( -4 \) and 6, so \( \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \). Perimeter: \( 10 + 2\sqrt{13} \approx 17.21 \). Image perimeter. \( A'B' \): from \( (4,-1) \) to \( (8,-1) \), length 4. \( A'C' \): from \( (4,-1) \) to \( (4,5) \), length 6. \( B'C' \): from \( (8,-1) \) to \( (4,5) \), differences \( -4 \) and 6, so \( 2\sqrt{13} \). Perimeter: \( 10 + 2\sqrt{13} \), identical. The perimeters had to match, since a translation preserves every distance and a perimeter is a sum of distances. Computing both is a check on the arithmetic rather than a discovery. \( B'(8,-1) \), \( C'(4,5) \); both perimeters \( 10 + 2\sqrt{13} \approx 17.21 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Reflect \( (4, 7) \) across the \( x \) axis.
   Show the full solution

   \( (4, -7) \)

2. Reflect \( (4, 7) \) across the \( y \) axis.
   Show the full solution

   \( (-4, 7) \)

3. Reflect \( (3, 8) \) across \( y = x \).
   Show the full solution

   Swap the coordinates. \( (8, 3) \)

4. How many lines of symmetry does an equilateral triangle have?
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   Three

5. Which points are fixed by a reflection?
   Show the full solution

   Exactly the points on the line of reflection

6. Reflect the triangle \( (1,2) \), \( (4,2) \), \( (1,6) \) across the \( y \) axis, and confirm the side lengths are unchanged.
   Show the full solution

   Images: \( (-1,2) \), \( (-4,2) \), \( (-1,6) \). Original sides: 3 horizontal, 4 vertical, and \( \sqrt{9 + 16} = 5 \) for the third. Image sides: from \( (-1,2) \) to \( (-4,2) \) is 3; from \( (-1,2) \) to \( (-1,6) \) is 4; from \( (-4,2) \) to \( (-1,6) \) is \( \sqrt{9 + 16} = 5 \). All unchanged. Orientation, however, is reversed: the original vertices in the order given run counterclockwise, and the images run clockwise. Images \( (-1,2) \), \( (-4,2) \), \( (-1,6) \); sides 3, 4, 5 in both

7. Reflect \( (5, -2) \) across \( y = -x \).
   Show the full solution

   The rule is \( (x, y) \to (-y, -x) \), so \( (5, -2) \to (2, -5) \). Check with the definition. The midpoint of the segment from \( (5,-2) \) to \( (2,-5) \) is \( \left( \dfrac{7}{2}, -\dfrac{7}{2} \right) \), which satisfies \( y = -x \). And the segment has slope \( \dfrac{-5 - (-2)}{2 - 5} = \dfrac{-3}{-3} = 1 \), while the line \( y = -x \) has slope \( -1 \), and \( 1 \times (-1) = -1 \), so they are perpendicular. Both conditions hold. \( (2, -5) \)

8. How many lines of symmetry does a regular hexagon have, and describe them.
   Show the full solution

   Six, which is the general result for a regular \( n \)-gon. They come in two families here, because 6 is even. Three pass through pairs of opposite vertices. Reflecting across one of these fixes two vertices and swaps the other four in pairs. Three pass through the midpoints of pairs of opposite sides. Reflecting across one of these fixes no vertices and swaps all six in pairs. For a regular polygon with an odd number of sides the picture differs: each line of symmetry passes through one vertex and the midpoint of the opposite side, so there is only one family. An equilateral triangle is the smallest case. Six: three through opposite vertices and three through opposite side midpoints

9. Explain why reflecting twice across the same line returns every point to its start.
   Show the full solution

   Take a point \( P \) not on the line \( \ell \), with image \( P' \). By the definition, \( \ell \) is the perpendicular bisector of \( \overline{PP'} \). Now reflect \( P' \) across \( \ell \). Its image is the point \( P'' \) such that \( \ell \) is the perpendicular bisector of \( \overline{P'P''} \). But \( P \) already has that property, and the perpendicular bisector condition determines the image uniquely: there is exactly one point on the perpendicular through \( P' \), at the same distance on the other side of \( \ell \). So \( P'' = P \). A point on \( \ell \) is fixed by the first reflection and therefore by the second, so it returns to itself trivially. Every point therefore comes back, and the composition is the identity. A reflection is its own inverse, which distinguishes it from a translation, where the inverse is a different translation, and from most rotations. The perpendicular bisector condition determines the image uniquely, so reflecting \( P' \) returns exactly \( P \)

10. A reflection maps \( (1, 4) \) to \( (7, 4) \). Find the line of reflection, and use it to find the image of \( (3, 9) \).
    Show the full solution

    Find the line. By the definition, the line of reflection is the perpendicular bisector of the segment joining the point to its image. Midpoint of \( (1,4) \) and \( (7,4) \): \( \left( \dfrac{1+7}{2}, \dfrac{4+4}{2} \right) = (4, 4) \). The segment is horizontal, so its perpendicular bisector is vertical, and it passes through \( (4,4) \). The line is \( x = 4 \). Find the image of \( (3,9) \). Reflecting across the vertical line \( x = 4 \) leaves \( y \) alone and sends \( x \) to the mirror position. The point \( (3,9) \) is 1 unit left of the line, so its image is 1 unit right: \( (5, 9) \). In rule form, reflection across \( x = c \) is \( (x, y) \to (2c - x, y) \), and here \( 2(4) - 3 = 5 \). The same rule checks the given pair: \( 2(4) - 1 = 7 \). Correct. Line \( x = 4 \); the image of \( (3,9) \) is \( (5,9) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Rotate \( (2, 5) \) by \( 90^\circ \) counterclockwise about the origin.
   Show the full solution

   \( (x,y) \to (-y, x) \). \( (-5, 2) \)

2. Rotate \( (2, 5) \) by \( 180^\circ \) about the origin.
   Show the full solution

   \( (-2, -5) \)

3. Rotate \( (2, 5) \) by \( 270^\circ \) counterclockwise about the origin.
   Show the full solution

   \( (x,y) \to (y, -x) \). \( (5, -2) \)

4. What is the smallest rotation mapping a regular octagon to itself?
   Show the full solution

   \( 360 \div 8 \). \( 45^\circ \)

5. Which point is fixed by a rotation?
   Show the full solution

   Only the center

6. Rotate the triangle \( (1,1) \), \( (4,1) \), \( (1,3) \) by \( 90^\circ \) counterclockwise about the origin, and confirm the side lengths.
   Show the full solution

   Applying \( (x,y) \to (-y,x) \): \( (1,1) \to (-1,1) \); \( (4,1) \to (-1,4) \); \( (1,3) \to (-3,1) \). Original sides: 3 horizontal, 2 vertical, and \( \sqrt{9 + 4} = \sqrt{13} \). Image sides: from \( (-1,1) \) to \( (-1,4) \) is 3; from \( (-1,1) \) to \( (-3,1) \) is 2; from \( (-1,4) \) to \( (-3,1) \) is \( \sqrt{4 + 9} = \sqrt{13} \). All three preserved, and the horizontal side became vertical, as a quarter turn requires. Images \( (-1,1) \), \( (-1,4) \), \( (-3,1) \)

7. What is the order of rotational symmetry of a parallelogram that is not a rectangle?
   Show the full solution

   Rotating a parallelogram \( 180^\circ \) about the intersection of its diagonals sends each vertex to the opposite one, and the figure lands on itself. That is one nontrivial rotation. A \( 90^\circ \) rotation does not work in general, since it would send a side onto an adjacent side, and in a non-rectangular parallelogram adjacent sides usually have different lengths and always meet at a non-right angle. So the rotations mapping the figure to itself are \( 180^\circ \) and \( 360^\circ \), giving order 2. A figure of order 2 is said to have point symmetry. Note that a general parallelogram has no lines of symmetry at all, so rotational and reflective symmetry are genuinely independent properties. Order 2, with a \( 180^\circ \) rotation about the diagonal intersection

8. A rotation of \( 90^\circ \) counterclockwise about the origin maps \( P \) to \( (6, -2) \). Find \( P \).
   Show the full solution

   The rotation rule is \( (x, y) \to (-y, x) \), so if \( P = (x, y) \) then \( (-y, x) = (6, -2) \). Matching coordinates: \( -y = 6 \) gives \( y = -6 \), and \( x = -2 \). So \( P = (-2, -6) \). Check by applying the rule forward: \( (-2, -6) \to (6, -2) \). Correct. An alternative route is to undo the rotation, which means rotating \( (6,-2) \) by \( 90^\circ \) clockwise using \( (x,y) \to (y,-x) \), giving \( (-2, -6) \). The same answer, and often quicker. \( P(-2, -6) \)

9. Explain why a \( 180^\circ \) rotation has the same rule clockwise and counterclockwise.
   Show the full solution

   Turning a half turn one way and a half turn the other way both bring a point to the position diametrically opposite the center, so they land in the same place. Concretely, going counterclockwise by \( 180^\circ \) and clockwise by \( 180^\circ \) differ by a full \( 360^\circ \) turn, which returns every point to where it started. Two rotations differing by a full turn are the same transformation. In coordinates, counterclockwise \( 180^\circ \) is \( (x,y) \to (-x,-y) \). Applying the counterclockwise \( 90^\circ \) rule twice gives \( (x,y) \to (-y,x) \to (-x,-y) \), and applying the clockwise \( 90^\circ \) rule twice gives \( (x,y) \to (y,-x) \to (-x,-y) \). Identical. This is the only rotation angle with this property. A \( 90^\circ \) turn and a \( 270^\circ \) turn go to different places, which is why direction has to be stated for every angle except \( 180^\circ \). The two differ by a full turn, which is the identity, so they are the same transformation

10. Rotate \( (5, 2) \) by \( 90^\circ \) counterclockwise about the point \( (1, 1) \) rather than the origin.
    Show the full solution

    The coordinate rules are stated for the origin, so the problem is solved by moving the center to the origin, rotating, and moving back. Step one: translate so the center goes to the origin. Subtract \( (1,1) \) from the point: \( (5 - 1, 2 - 1) = (4, 1) \). Step two: rotate about the origin. Apply \( (x,y) \to (-y,x) \): \( (4,1) \to (-1, 4) \). Step three: translate back. Add \( (1,1) \): \( (-1 + 1, 4 + 1) = (0, 5) \). Check the two conditions of a rotation. Distance from the center \( (1,1) \) to the original \( (5,2) \): differences 4 and 1, so \( \sqrt{16 + 1} = \sqrt{17} \). Distance from \( (1,1) \) to the image \( (0,5) \): differences \( -1 \) and 4, so \( \sqrt{1 + 16} = \sqrt{17} \). Equal. Slopes from the center: to \( (5,2) \) the slope is \( \dfrac{1}{4} \); to \( (0,5) \) it is \( \dfrac{4}{-1} = -4 \). Product: \( \dfrac{1}{4} \times (-4) = -1 \), so the two radii are perpendicular and the turn is a right angle. Both conditions hold, confirming the image. This translate-rotate-translate pattern handles any center, and lesson 4.5 uses the same idea for compositions generally. \( (0, 5) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Does order matter in a composition?
   Show the full solution

   Usually yes

2. Two reflections in parallel lines compose to what?
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   A translation

3. Two reflections in intersecting lines compose to what?
   Show the full solution

   A rotation about their intersection

4. Is a composition of rigid motions rigid?
   Show the full solution

   Yes

5. What is a glide reflection?
   Show the full solution

   A reflection followed by a translation parallel to the line of reflection

6. Apply to \( (3, -1) \): reflect across the \( y \) axis, then translate by \( \langle 2, 5 \rangle \).
   Show the full solution

   First the reflection: \( (3,-1) \to (-3, -1) \). Then the translation: \( (-3 + 2, -1 + 5) = (-1, 4) \). \( (-1, 4) \)

7. Apply the same two transformations in the opposite order and compare.
   Show the full solution

   First the translation: \( (3,-1) \to (5, 4) \). Then the reflection across the \( y \) axis: \( (5,4) \to (-5, 4) \). The previous problem gave \( (-1, 4) \); this order gives \( (-5, 4) \). Different. The reason: the translation moves the point 2 units right, and reflecting afterward converts that into 2 units further left of the axis, whereas reflecting first puts the point on the left and then moves it 2 units back toward the axis. The horizontal component is affected by the reflection and the order therefore matters. The vertical component is unaffected, which is why both answers have \( y = 4 \). \( (-5, 4) \), different from \( (-1, 4) \)

8. Reflect \( (4, 2) \) across \( y = 1 \), then across \( y = 5 \). Identify the composition.
   Show the full solution

   Reflection across the horizontal line \( y = c \) is \( (x,y) \to (x, 2c - y) \). Across \( y = 1 \): \( (4, 2) \to (4, 2 - 2) = (4, 0) \). Across \( y = 5 \): \( (4, 0) \to (4, 10 - 0) = (4, 10) \). Net effect: \( (4,2) \to (4,10) \), a move of 8 units up with no horizontal change. That is a translation by \( \langle 0, 8 \rangle \). Checking against the rule: the lines are parallel and 4 units apart, and \( 2 \times 4 = 8 \), in the direction from the first line to the second, which is upward. Confirmed. A translation by \( \langle 0, 8 \rangle \)

9. Explain why two reflections in parallel lines give a translation of twice the distance between them.
   Show the full solution

   Set the two vertical lines at \( x = p \) and \( x = q \), with \( q \gt p \), so the distance between them is \( d = q - p \). Reflecting across \( x = p \) sends a point \( (x, y) \) to \( (2p - x, y) \). Reflecting that across \( x = q \) sends it to \( (2q - (2p - x), \; y) = (2q - 2p + x, \; y) = (x + 2(q - p), \; y) \). So the net rule is \( (x, y) \to (x + 2d, \; y) \), a translation of \( 2d \) horizontally, which is perpendicular to the vertical lines. The \( y \) coordinate never changed, and the result does not depend on \( x \), which is what makes it a translation rather than something more complicated. Geometrically: the first reflection moves a point to the far side of the first line, and the second moves it to the far side of the second. The two moves are in the same direction and their combined effect is the full width of the strip counted twice. The direction matters: reflecting in \( x = q \) first and then \( x = p \) gives a translation of \( 2d \) in the opposite direction, which is why the order changes the answer. The algebra gives \( x \to x + 2(q-p) \); both reflections push the point the same way, across the strip and then across it again

10. A glide reflection reflects across the \( x \) axis and translates by \( \langle 3, 0 \rangle \). Apply it twice to \( (1, 4) \) and describe the result.
    Show the full solution

    First application. Reflect: \( (1,4) \to (1, -4) \). Translate: \( (1 + 3, -4) = (4, -4) \). Second application. Reflect: \( (4,-4) \to (4, 4) \). Translate: \( (4 + 3, 4) = (7, 4) \). The result. The point went from \( (1,4) \) to \( (7,4) \), a translation of 6 units right with no vertical change. Applying a glide reflection twice gives a pure translation, of twice the glide vector. The reason is that the two reflections cancel: reflecting across the same line twice is the identity, as lesson 4.3 established. What survives is the two translations, which add. This works because the translation is parallel to the line of reflection, so the two parts do not interfere; that parallelism is part of the definition of a glide reflection. Note also that a glide reflection reverses orientation, since it contains one reflection, and applying it twice restores orientation, consistent with the result being a translation. \( (7,4) \), a translation by \( \langle 6, 0 \rangle \); the two reflections cancel and the translations add

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. How many lines of symmetry does a regular pentagon have?
   Show the full solution

   Five

2. What is the order of rotational symmetry of a regular pentagon?
   Show the full solution

   5

3. How many lines of symmetry does a circle have?
   Show the full solution

   Every line through the center works. Infinitely many

4. Does a general parallelogram have line symmetry?
   Show the full solution

   No

5. What order of rotational symmetry means point symmetry?
   Show the full solution

   Order 2

6. Describe all the symmetry of a rhombus that is not a square.
   Show the full solution

   Lines of symmetry: two, along the diagonals. Reflecting across a diagonal swaps the two triangles it creates, and they are congruent because all four sides of a rhombus are congruent. The lines through opposite side midpoints fail, unlike in a rectangle, because the sides are not perpendicular to those lines. Rotational symmetry: order 2, by a half turn about the intersection of the diagonals. A quarter turn fails because the two diagonals of a non-square rhombus have different lengths, so a quarter turn would send the long diagonal onto the short one. Comparing with the rectangle: both have two lines of symmetry and order 2, but the lines are in different places, along the diagonals for a rhombus and through side midpoints for a rectangle. The square is the figure that has both sets, giving four. Two lines of symmetry along the diagonals, rotational symmetry of order 2

7. A figure has rotational symmetry of order 3. What is its smallest angle of rotational symmetry?
   Show the full solution

   \( \dfrac{360^\circ}{3} = 120^\circ \). The full set of rotations mapping it to itself is \( 120^\circ \), \( 240^\circ \) and \( 360^\circ \), which is three of them, matching the order. An equilateral triangle is the standard example, and so is a three-bladed propeller shape, which has order 3 without having any line symmetry at all. \( 120^\circ \)

8. Explain why a diagonal of a non-square rectangle is not a line of symmetry.
   Show the full solution

   Take a rectangle with vertices \( A \), \( B \), \( C \), \( D \) in order, with \( AB = 5 \) and \( BC = 3 \), and consider the diagonal \( \overline{AC} \). A reflection across \( \overline{AC} \) fixes \( A \) and \( C \), since they lie on the line. It must therefore send \( B \) to \( D \), the only other vertex available. But a reflection preserves distance, so it would need \( AB = AD \). In this rectangle \( AB = 5 \) and \( AD = 3 \), which are different. The reflection cannot do it, so the diagonal is not a line of symmetry. The argument also shows exactly when it does work: if \( AB = AD \), the rectangle is a square, and then the diagonals are lines of symmetry. That is why a square has four and a rectangle two. The general lesson is the one the method states: test a vertex rather than judging by eye. A diagonal divides a rectangle into two congruent triangles, which makes it look like a line of symmetry, but the two triangles are related by a half turn rather than by a reflection. The reflection would fix \( A \) and send \( B \) to \( D \), requiring \( AB = AD \), which holds only for a square

9. A regular polygon has rotational symmetry with smallest angle \( 24^\circ \). How many sides does it have, and how many lines of symmetry?
   Show the full solution

   The smallest angle of rotational symmetry for a regular \( n \)-gon is \( \dfrac{360^\circ}{n} \), so \( \dfrac{360}{n} = 24 \), giving \( n = \dfrac{360}{24} = 15 \). A regular 15-gon therefore, with rotational symmetry of order 15. A regular \( n \)-gon has \( n \) lines of symmetry, so it has 15 lines of symmetry. Since 15 is odd, each line passes through one vertex and the midpoint of the opposite side, as with an equilateral triangle or a regular pentagon. Even-sided regular polygons have two families instead. 15 sides and 15 lines of symmetry

10. Give a figure with line symmetry but no rotational symmetry, and one with rotational symmetry but no line symmetry, and explain both.
    Show the full solution

    Line symmetry without rotational symmetry: an isosceles trapezoid. The vertical line through the midpoints of the two parallel sides is a line of symmetry: reflecting across it swaps the two congruent legs and swaps the base angles at each base, so the figure lands on itself. No rotation works. A half turn would send the shorter parallel side to the position of the longer one, and those have different lengths, which a rigid motion cannot do. Any smaller angle would not even send a vertex to a vertex. A non-equilateral isosceles triangle is another example, with one line of symmetry and no rotational symmetry. Rotational symmetry without line symmetry: a general parallelogram. A half turn about the intersection of the diagonals maps each vertex to the opposite one and the figure to itself, giving order 2. No reflection works. Reflecting across a diagonal would need the two sides meeting at a vertex on that diagonal to be congruent, which they are not in a general parallelogram. Reflecting across a line through opposite side midpoints would need the sides to be perpendicular to that line, which they are not unless the figure is a rectangle. A pinwheel or propeller shape is another example, with high rotational symmetry and no lines at all. Isosceles trapezoid has one line and no rotational symmetry; a general parallelogram has order 2 rotational symmetry and no lines

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Define congruent using rigid motions.
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   Two figures are congruent when a sequence of rigid motions maps one onto the other

2. If \( \triangle ABC \cong \triangle DEF \), which side corresponds to \( \overline{AB} \)?
   Show the full solution

   \( \overline{DE} \)

3. What does reversed orientation indicate about the sequence?
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   It includes a reflection

4. How many parts does a triangle have?
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   Six: three sides and three angles

5. If a rigid motion maps \( P \) to \( Q \), what is true of any distance from \( P \)?
   Show the full solution

   It is preserved

6. \( \triangle ABC \) has \( A(0,0) \), \( B(3,0) \), \( C(0,4) \). \( \triangle PQR \) has \( P(5,2) \), \( Q(8,2) \), \( R(5,6) \). Find the rigid motion and write the congruence statement.
   Show the full solution

   Compare coordinates: \( A(0,0) \to P(5,2) \) adds 5 and 2. Check the others: \( B(3,0) + (5,2) = (8,2) = Q \). Correct. \( C(0,4) + (5,2) = (5,6) = R \). Correct. The transformation is a translation by \( \langle 5, 2 \rangle \). Since the correspondence is \( A \to P \), \( B \to Q \), \( C \to R \), the statement is \( \triangle ABC \cong \triangle PQR \). Orientation is unchanged, consistent with a translation containing no reflection. Translation by \( \langle 5,2 \rangle \); \( \triangle ABC \cong \triangle PQR \)

7. \( \triangle ABC \) has \( A(1,2) \), \( B(4,2) \), \( C(1,6) \). \( \triangle XYZ \) has \( X(-1,2) \), \( Y(-4,2) \), \( Z(-1,6) \). Identify the motion.
   Show the full solution

   Each image has the same \( y \) and the opposite \( x \): \( (1,2) \to (-1,2) \), \( (4,2) \to (-4,2) \), \( (1,6) \to (-1,6) \). The rule is \( (x,y) \to (-x, y) \), reflection across the \( y \) axis. Verify all three vertices land correctly: they do. Congruence statement: \( \triangle ABC \cong \triangle XYZ \). Orientation is reversed, as a reflection requires, which is a useful independent check on the identification. Reflection across the \( y \) axis

8. Why is "same size and shape" not adequate as a definition of congruence?
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   It is a description of what congruence feels like rather than a statement a proof can use. "Size" and "shape" are not themselves defined, so the phrase explains one undefined idea with two more. More practically, it gives no procedure. Faced with two figures, there is nothing to check and no way to settle a disagreement. The rigid motion definition gives an exact test: either a sequence of translations, reflections and rotations carries one figure onto the other, or it does not, and exhibiting the sequence settles it. The definition is also what makes the triangle criteria provable. SSS is not an additional assumption; it is a theorem, derived by showing that three pairs of congruent sides are enough to build the sequence. From "same size and shape" no such derivation is possible. Finally, the rigid motion definition extends to figures other than triangles and to three dimensions without change, which a shape-based description does not do cleanly. It defines one vague idea with two others, gives no test to apply, and cannot support a proof of the triangle criteria

9. \( \triangle ABC \) with \( A(2,1) \), \( B(5,1) \), \( C(2,3) \) maps to \( \triangle DEF \) with \( D(-1,2) \), \( E(-1,5) \), \( F(-3,2) \). Find the sequence.
   Show the full solution

   Look for a rotation first, since the horizontal side became vertical. Try \( 90^\circ \) counterclockwise about the origin, \( (x,y) \to (-y, x) \): \( A(2,1) \to (-1, 2) = D \). Correct. \( B(5,1) \to (-1, 5) = E \). Correct. \( C(2,3) \to (-3, 2) = F \). Correct. All three vertices land on target with a single transformation, so the sequence is one rotation of \( 90^\circ \) counterclockwise about the origin. Checking orientation: \( A, B, C \) runs counterclockwise and \( D, E, F \) also runs counterclockwise, consistent with a rotation, which preserves orientation. Had orientation flipped, a rotation alone could not have been the answer. Congruence statement: \( \triangle ABC \cong \triangle DEF \). A single rotation of \( 90^\circ \) counterclockwise about the origin

10. Explain why SSS gives congruence, using the rigid motion definition.
    Show the full solution

    Suppose \( \triangle ABC \) and \( \triangle DEF \) have \( AB = DE \), \( BC = EF \) and \( AC = DF \). Build the sequence in three moves. Move one: translate. Translate \( \triangle ABC \) by the vector from \( A \) to \( D \). Now \( A \) coincides with \( D \), and the triangle is otherwise unchanged, since a translation is rigid. Move two: rotate. Rotate about \( D \) until the ray \( \overrightarrow{AB} \) lies along the ray \( \overrightarrow{DE} \). This is possible because a rotation can turn a ray to any direction about its endpoint. Since \( AB = DE \) and the rotation preserved that length, the point \( B \) now coincides exactly with \( E \), not merely lying in the same direction. Move three: reflect if necessary. Two vertices are now in place, so the third, currently at some point \( C' \), must satisfy \( DC' = AC = DF \) and \( EC' = BC = EF \). Exactly two points in the plane are at those two distances from \( D \) and \( E \), and they are reflections of each other across line \( DE \). One of them is \( F \). If \( C' \) is already \( F \), stop; if not, reflect across \( \overleftrightarrow{DE} \), which fixes \( D \) and \( E \) and sends \( C' \) to \( F \). The sequence, at most a translation then a rotation then a reflection, maps \( \triangle ABC \) exactly onto \( \triangle DEF \). By the definition, the triangles are congruent. Note what the argument used: only the three side lengths, and the fact that two circles centered at \( D \) and \( E \) meet in at most two points. Nothing about the angles was assumed, which is why they end up determined rather than needing to be given. Translate a vertex into place, rotate a side into place, and reflect if needed; the three given lengths leave the third vertex only two possible positions, which a reflection resolves

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. If \( \triangle ABC \cong \triangle DEF \), which angle corresponds to \( \angle B \)?
   Show the full solution

   \( \angle E \)

2. Which side corresponds to \( \overline{BC} \)?
   Show the full solution

   \( \overline{EF} \)

3. How many pairs of congruent parts does a triangle congruence give?
   Show the full solution

   Six

4. If \( \triangle PQR \cong \triangle XYZ \) and \( PQ = 9 \), find \( XY \).
   Show the full solution

   9

5. If \( \triangle LMN \cong \triangle RST \) and \( m\angle N = 55^\circ \), find \( m\angle T \).
   Show the full solution

   \( 55^\circ \)

6. \( \triangle ABC \cong \triangle FDE \). List all six pairs of congruent parts.
   Show the full solution

   The correspondence is \( A \to F \), \( B \to D \), \( C \to E \), read in position. Angles: \( \angle A \cong \angle F \), \( \angle B \cong \angle D \), \( \angle C \cong \angle E \). Sides: \( \overline{AB} \cong \overline{FD} \), \( \overline{BC} \cong \overline{DE} \), \( \overline{AC} \cong \overline{FE} \). The letters of the second triangle are deliberately not in alphabetical order, which is the whole point: the order records the correspondence rather than a naming convention. Three angle pairs and three side pairs as listed

7. \( \triangle ABC \cong \triangle DEF \) with \( m\angle A = 50^\circ \), \( m\angle E = 60^\circ \). Find all six angle measures.
   Show the full solution

   From the correspondence, \( \angle A \cong \angle D \) and \( \angle B \cong \angle E \), so \( m\angle D = 50^\circ \) and \( m\angle B = 60^\circ \). In \( \triangle ABC \), the angle sum gives \( m\angle C = 180 - 50 - 60 = 70^\circ \), and therefore \( m\angle F = 70^\circ \) by correspondence. All six: \( \angle A = \angle D = 50^\circ \), \( \angle B = \angle E = 60^\circ \), \( \angle C = \angle F = 70^\circ \). Check: each triangle's angles sum to \( 50 + 60 + 70 = 180^\circ \). Correct. \( 50^\circ \), \( 60^\circ \), \( 70^\circ \) in each triangle

8. Two triangles are congruent with \( \overline{AB} \cong \overline{QR} \), \( \overline{BC} \cong \overline{RP} \), \( \overline{CA} \cong \overline{PQ} \). Write the statement.
   Show the full solution

   Work out the vertex correspondence from the side markings. \( \overline{AB} \cong \overline{QR} \) pairs \( \{A, B\} \) with \( \{Q, R\} \). \( \overline{BC} \cong \overline{RP} \) pairs \( \{B, C\} \) with \( \{R, P\} \). The letter \( B \) appears in both, and \( R \) is the letter common to both image pairs, so \( B \to R \). Then from the first pairing, \( A \to Q \), and from the second, \( C \to P \). Check against the third marking: \( \overline{CA} \) should correspond to \( \overline{PQ} \), and indeed \( C \to P \) and \( A \to Q \). Consistent. \( \triangle ABC \cong \triangle QRP \)

9. Explain why \( \triangle ABC \cong \triangle DEF \) and \( \triangle ABC \cong \triangle EDF \) cannot both be true unless the triangle is isosceles.
   Show the full solution

   The first statement gives \( \overline{AB} \cong \overline{DE} \) and \( \overline{AC} \cong \overline{DF} \). The second gives \( \overline{AB} \cong \overline{ED} \), which is the same segment as \( \overline{DE} \), so that is consistent, but it also gives \( \overline{AC} \cong \overline{EF} \) and \( \overline{BC} \cong \overline{DF} \). Combining, \( \overline{AC} \cong \overline{DF} \) from the first and \( \overline{BC} \cong \overline{DF} \) from the second give \( \overline{AC} \cong \overline{BC} \) by the transitive property. So \( \triangle ABC \) has two congruent sides and is isosceles. If it is scalene, both statements cannot hold. Both can hold when the triangle is isosceles, because then two different correspondences genuinely work, and both statements are true. That is the exception, not the rule, and it is why a congruence statement has to be justified by the correspondence rather than written down and hoped for. Together they force \( AC = BC \), so the triangle must be isosceles

10. \( \triangle ABC \cong \triangle DEF \). Given \( AB = 2x + 3 \), \( DE = 11 \), \( m\angle C = 5y \), \( m\angle F = 40^\circ \). Find \( x \), \( y \) and the perimeter of \( \triangle DEF \) if \( BC = 9 \) and \( AC = 7 \).
    Show the full solution

    Find \( x \). Corresponding sides are congruent, so \( \overline{AB} \cong \overline{DE} \) gives \( AB = DE \): \( 2x + 3 = 11 \), so \( 2x = 8 \) and \( x = 4 \). Then \( AB = 11 \), matching. Find \( y \). Corresponding angles give \( m\angle C = m\angle F \): \( 5y = 40 \), so \( y = 8 \). Then \( m\angle C = 40^\circ \), matching. Find the perimeter of \( \triangle DEF \). Each side of \( \triangle DEF \) equals its corresponding side in \( \triangle ABC \): \( DE = AB = 11 \), \( EF = BC = 9 \), \( DF = AC = 7 \). Perimeter: \( 11 + 9 + 7 = 27 \). Note that the perimeter of \( \triangle ABC \) is also 27, which it must be: congruent figures have equal perimeters because every corresponding side is congruent. Computing both is a check rather than extra work. Note also the triangle inequality holds, \( 7 + 9 = 16 \gt 11 \), so these lengths can form a triangle. Unit 6 makes that test explicit. \( x = 4 \), \( y = 8 \), perimeter 27

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does SSS require?
   Show the full solution

   Three pairs of congruent sides

2. What does SAS require?
   Show the full solution

   Two pairs of congruent sides and the included angles congruent

3. In \( \triangle ABC \), which angle is included between \( \overline{AB} \) and \( \overline{BC} \)?
   Show the full solution

   The shared vertex is \( B \). \( \angle B \)

4. What property justifies \( \overline{XY} \cong \overline{XY} \)?
   Show the full solution

   Reflexive property of congruence

5. Two triangles have \( \overline{AB} \cong \overline{DE} \), \( \overline{AC} \cong \overline{DF} \), \( \angle A \cong \angle D \). Which criterion applies?
   Show the full solution

   \( \angle A \) is between \( \overline{AB} \) and \( \overline{AC} \). SAS

6. Given \( \overline{AB} \cong \overline{AD} \) and \( \overline{CB} \cong \overline{CD} \), prove \( \triangle ABC \cong \triangle ADC \).
   Show the full solution

   The two triangles share \( \overline{AC} \), which supplies the third pair. 1. \( \overline{AB} \cong \overline{AD} \) and \( \overline{CB} \cong \overline{CD} \). Reason: given. 2. \( \overline{AC} \cong \overline{AC} \). Reason: reflexive property of congruence. 3. \( \triangle ABC \cong \triangle ADC \). Reason: SSS. The figure here is a kite, and this proof is the one unit 7 uses to establish the kite properties. Proved by SSS

7. Given \( \overline{AB} \cong \overline{CB} \) and \( \overline{BD} \) bisects \( \angle ABC \), prove \( \triangle ABD \cong \triangle CBD \).
   Show the full solution

   1. \( \overline{AB} \cong \overline{CB} \) and \( \overline{BD} \) bisects \( \angle ABC \). Reason: given. 2. \( \angle ABD \cong \angle CBD \). Reason: definition of an angle bisector. 3. \( \overline{BD} \cong \overline{BD} \). Reason: reflexive property of congruence. 4. \( \triangle ABD \cong \triangle CBD \). Reason: SAS. Check that the angle is included: in \( \triangle ABD \) the sides are \( \overline{AB} \) and \( \overline{BD} \), and \( \angle ABD \) sits between them at vertex \( B \). Included, so SAS is legitimate. Proved by SAS

8. Two triangles have \( \overline{PQ} \cong \overline{XY} \), \( \overline{QR} \cong \overline{YZ} \) and \( \angle P \cong \angle X \). Can you conclude congruence?
   Show the full solution

   Check whether the angle is included. In \( \triangle PQR \) the two given sides are \( \overline{PQ} \) and \( \overline{QR} \), which meet at \( Q \). So the included angle is \( \angle Q \), not \( \angle P \). The given angle \( \angle P \) is opposite \( \overline{QR} \), so the configuration is side, side, non-included angle, which is SSA. SSA is not a valid criterion, as lesson 5.3 demonstrates with two genuinely different triangles sharing the same SSA data. So no conclusion follows. What would fix it: being given \( \angle Q \cong \angle Y \) instead, which would make it SAS, or being given the third side, which would make it SSS. No; the angle is not included, so this is SSA and proves nothing

9. Explain why the reflexive property is needed so often in this unit.
   Show the full solution

   Most proof diagrams put the two triangles next to each other sharing a side or an angle, because that is how congruent triangles arise naturally inside a larger figure: a diagonal cutting a quadrilateral, a segment drawn from a vertex, two triangles meeting at a point. In such a figure, the shared part is genuinely one object appearing in both triangles. But a congruence criterion needs three pairs, and a pair means two parts related by a congruence statement. The reflexive property supplies exactly that: the shared side is congruent to itself, so it counts as a pair. Omitting the line does not make the proof wrong in substance, but it leaves only two pairs on the page, which is not enough to invoke any criterion. A reader checking the proof line by line finds a criterion cited with insufficient support. The habit to build: as soon as two triangles are identified, look for what they share and write the reflexive line immediately, before counting up to three. Overlapping triangles share a part, and the reflexive property is what turns that one shared object into the third congruent pair a criterion needs

10. In quadrilateral \( ABCD \), the diagonals \( \overline{AC} \) and \( \overline{BD} \) bisect each other at \( E \). Prove \( \triangle AEB \cong \triangle CED \) and then that \( \overline{AB} \cong \overline{CD} \).
    Show the full solution

    1. \( \overline{AC} \) and \( \overline{BD} \) bisect each other at \( E \). Reason: given. 2. \( E \) is the midpoint of \( \overline{AC} \) and of \( \overline{BD} \). Reason: definition of a segment bisector. 3. \( \overline{AE} \cong \overline{EC} \). Reason: definition of midpoint. 4. \( \overline{BE} \cong \overline{ED} \). Reason: definition of midpoint. 5. \( \angle AEB \cong \angle CED \). Reason: vertical angles theorem. 6. \( \triangle AEB \cong \triangle CED \). Reason: SAS. 7. \( \overline{AB} \cong \overline{CD} \). Reason: corresponding parts of congruent triangles are congruent. Checking the included angle at line 6. In \( \triangle AEB \) the sides in hand are \( \overline{AE} \) and \( \overline{BE} \), meeting at \( E \), so \( \angle AEB \) is included. Legitimate. What this proves. A quadrilateral whose diagonals bisect each other has a pair of opposite sides congruent, and the same argument on the other pair of triangles gives \( \overline{AD} \cong \overline{CB} \). That is most of the proof that such a quadrilateral is a parallelogram, which unit 7 completes. Proved by SAS, with the side congruence following by CPCTC

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does ASA require?
   Show the full solution

   Two angles and the included side

2. What does AAS require?
   Show the full solution

   Two angles and a non-included side

3. Is SSA a valid criterion?
   Show the full solution

   No

4. Is AAA a valid criterion for congruence?
   Show the full solution

   It gives similarity only. No

5. In \( \triangle ABC \), which side is included between \( \angle A \) and \( \angle C \)?
   Show the full solution

   \( \overline{AC} \)

6. Two triangles have \( \angle A \cong \angle D \), \( \angle C \cong \angle F \), \( \overline{AC} \cong \overline{DF} \). Which criterion applies?
   Show the full solution

   The side \( \overline{AC} \) joins the two named angles \( \angle A \) and \( \angle C \), so it is included. ASA

7. Two triangles have \( \angle A \cong \angle D \), \( \angle B \cong \angle E \), \( \overline{AC} \cong \overline{DF} \). Which criterion applies?
   Show the full solution

   The named angles are \( \angle A \) and \( \angle B \), so the included side would be \( \overline{AB} \). The given side \( \overline{AC} \) is not it. AAS

8. Given \( \overline{AB} \parallel \overline{CD} \) and \( M \) is the midpoint of \( \overline{AD} \), prove \( \triangle ABM \cong \triangle DCM \).
   Show the full solution

   1. \( \overline{AB} \parallel \overline{CD} \); \( M \) is the midpoint of \( \overline{AD} \). Reason: given. 2. \( \angle BAM \cong \angle CDM \). Reason: alternate interior angles theorem, with \( \overline{AD} \) as the transversal. 3. \( \overline{AM} \cong \overline{DM} \). Reason: definition of midpoint. 4. \( \angle AMB \cong \angle DMC \). Reason: vertical angles theorem. 5. \( \triangle ABM \cong \triangle DCM \). Reason: ASA. Check that the side is included: \( \overline{AM} \) joins \( \angle A \) and \( \angle M \) in the first triangle, which are the two named angles. Included, so ASA is correct. Proved by ASA

9. Explain in your own words why the word "included" matters so much in SAS.
   Show the full solution

   An included angle locks the two sides into a fixed relative position. Once the angle at a vertex is fixed and the two sides leaving that vertex have fixed lengths, the two far endpoints are determined, and so is the segment joining them. Nothing is left free, which is why the triangle is determined. A non-included angle does not do this. The angle sits at the far end of one of the sides, and the second side, of known length, can swing to meet the remaining side of the angle in two different places. That freedom is the ambiguous case, and it is what allows two non-congruent triangles to share the same SSA data. The worked example makes it concrete: with \( \angle A = 30^\circ \), \( AB = 7 \) and \( BC = 5 \), the arc of radius 5 about \( B \) crosses the other side of the angle twice, giving third sides of about 9.63 and 2.49. The general lesson: a criterion works when the given parts leave no freedom, and the word included is what removes the freedom in SAS. An included angle fixes the two sides relative to each other, leaving nothing free; a non-included angle allows the second side to reach the third in two places

10. Given \( \angle B \) and \( \angle E \) are right angles, \( \angle A \cong \angle D \), and \( \overline{AC} \cong \overline{DF} \), prove \( \triangle ABC \cong \triangle DEF \). Name the criterion and explain why a second one also applies.
    Show the full solution

    1. \( \angle B \) and \( \angle E \) are right angles; \( \angle A \cong \angle D \); \( \overline{AC} \cong \overline{DF} \). Reason: given. 2. \( \angle B \cong \angle E \). Reason: right angle congruence theorem. 3. \( \triangle ABC \cong \triangle DEF \). Reason: AAS. Checking the criterion. The two angle pairs are \( \angle A \) with \( \angle D \) and \( \angle B \) with \( \angle E \). The side between them would be \( \overline{AB} \), and the given side is \( \overline{AC} \), which is not it. So this is AAS, not ASA. Why HL also applies. The triangles are right triangles, since \( \angle B \) and \( \angle E \) are right angles. The side \( \overline{AC} \) is opposite \( \angle B \), so it is the hypotenuse of \( \triangle ABC \), and likewise \( \overline{DF} \) is the hypotenuse of \( \triangle DEF \). The given congruence is therefore a pair of congruent hypotenuses. HL needs a leg as well, however, and none is given here, so HL cannot be used as stated. AAS is the right citation. Had a leg been given instead of the angle at \( A \), HL would apply and AAS would not, which is the subject of lesson 5.4. Proved by AAS; the triangles are right, but HL needs a leg that is not given

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does HL require?
   Show the full solution

   Congruent hypotenuses and one pair of congruent legs, in right triangles

2. Which side of a right triangle is the hypotenuse?
   Show the full solution

   The one opposite the right angle

3. Can HL be used on triangles that are not right triangles?
   Show the full solution

   No

4. Two right triangles have congruent legs and congruent legs. Which criterion applies?
   Show the full solution

   The right angles are included between the legs. SAS

5. Name all five congruence criteria.
   Show the full solution

   SSS, SAS, ASA, AAS, HL

6. A right triangle has hypotenuse 25 and a leg of 7. Find the other leg.
   Show the full solution

   By the Pythagorean theorem, \( 7^2 + b^2 = 25^2 \), so \( b^2 = 625 - 49 = 576 \) and \( b = 24 \). A single positive value, which is exactly why HL works: the third side is not free. 24

7. Given \( \overline{PQ} \perp \overline{QR} \), \( \overline{ST} \perp \overline{TU} \), \( \overline{PR} \cong \overline{SU} \) and \( \overline{QR} \cong \overline{TU} \), prove \( \triangle PQR \cong \triangle STU \).
   Show the full solution

   1. \( \overline{PQ} \perp \overline{QR} \), \( \overline{ST} \perp \overline{TU} \), \( \overline{PR} \cong \overline{SU} \), \( \overline{QR} \cong \overline{TU} \). Reason: given. 2. \( \angle Q \) and \( \angle T \) are right angles. Reason: definition of perpendicular lines. 3. \( \triangle PQR \) and \( \triangle STU \) are right triangles. Reason: definition of a right triangle. 4. \( \overline{PR} \) and \( \overline{SU} \) are the hypotenuses. Reason: each is opposite the right angle. 5. \( \triangle PQR \cong \triangle STU \). Reason: HL. Proved by HL

8. Explain why HL does not contradict the failure of SSA.
   Show the full solution

   SSA fails because two triangles can satisfy the same data, one with an acute angle and one with an obtuse angle opposite the given side. Both are genuinely possible when nothing rules the obtuse case out. In a right triangle, the obtuse case is impossible. One angle already measures \( 90^\circ \), and the angle sum leaves only \( 90^\circ \) to be shared between the other two, so both must be acute. The second of the two candidate triangles cannot exist. So HL is not an exception to the SSA failure; it is the special situation in which the ambiguity that causes the failure cannot arise. The hypothesis "right triangle" is doing exactly that work, which is why a proof must establish it before citing HL. The algebraic version says the same thing: \( b = \sqrt{c^2 - a^2} \) has one positive solution, whereas the general SSA configuration has no such equation and leaves the third side genuinely undetermined. The right angle forces both other angles to be acute, eliminating the second triangle that makes SSA ambiguous

9. In an isosceles triangle \( ABC \) with \( \overline{AB} \cong \overline{AC} \), the altitude from \( A \) meets \( \overline{BC} \) at \( D \). Prove \( D \) is the midpoint of \( \overline{BC} \).
   Show the full solution

   1. \( \overline{AB} \cong \overline{AC} \); \( \overline{AD} \) is an altitude to \( \overline{BC} \). Reason: given. 2. \( \overline{AD} \perp \overline{BC} \). Reason: definition of an altitude. 3. \( \angle ADB \) and \( \angle ADC \) are right angles. Reason: definition of perpendicular lines. 4. \( \triangle ABD \) and \( \triangle ACD \) are right triangles. Reason: definition of a right triangle. 5. \( \overline{AD} \cong \overline{AD} \). Reason: reflexive property of congruence. 6. \( \triangle ABD \cong \triangle ACD \). Reason: HL, with \( \overline{AB} \) and \( \overline{AC} \) the hypotenuses. 7. \( \overline{BD} \cong \overline{DC} \). Reason: corresponding parts of congruent triangles are congruent. 8. \( D \) is the midpoint of \( \overline{BC} \). Reason: definition of midpoint, since \( D \) lies on \( \overline{BC} \) and divides it into two congruent segments. This proves a useful fact: in an isosceles triangle, the altitude to the base is also the median to the base. Lesson 5.7 shows it is the angle bisector as well. Proved by HL and CPCTC

10. Two right triangles have a congruent leg and a congruent acute angle. Determine whether they must be congruent, considering both possible positions of the angle.
    Show the full solution

    The acute angle can be either adjacent to the given leg or opposite it, and both cases turn out to give congruence, for slightly different reasons. Case one: the acute angle is adjacent to the given leg. Then the triangles share the right angle, the given leg, and the acute angle at the other end of that leg. The leg is between the right angle and the acute angle, so this is ASA. Congruent. Case two: the acute angle is opposite the given leg. Then the triangles share the right angle, the acute angle, and a non-included side. That is AAS. Congruent. Why both work. Knowing one acute angle in a right triangle determines the other, since the two must sum to \( 90^\circ \). So a right triangle with one known acute angle has all three angles known, and any one side then fixes the size completely. This is worth remembering as a practical shortcut: for right triangles, any one side plus any one acute angle gives congruence, and the only question is whether to cite ASA or AAS. Congruent in both cases, by ASA when the angle is adjacent and AAS when it is opposite

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does CPCTC stand for?
   Show the full solution

   Corresponding parts of congruent triangles are congruent

2. When in a proof may CPCTC be used?
   Show the full solution

   Only after the triangles have been proved congruent

3. \( \triangle ABC \cong \triangle DEF \). What does CPCTC give about \( \overline{AC} \)?
   Show the full solution

   \( \overline{AC} \cong \overline{DF} \)

4. Can CPCTC supply one of the three pairs for a criterion?
   Show the full solution

   No; that would be circular

5. What usually comes after CPCTC in a proof?
   Show the full solution

   A definition converting the congruent parts into the goal's language

6. Given \( M \) is the midpoint of \( \overline{AB} \) and \( \overline{CM} \perp \overline{AB} \), prove \( \overline{CA} \cong \overline{CB} \).
   Show the full solution

   1. \( M \) is the midpoint of \( \overline{AB} \); \( \overline{CM} \perp \overline{AB} \). Reason: given. 2. \( \overline{AM} \cong \overline{MB} \). Reason: definition of midpoint. 3. \( \angle CMA \) and \( \angle CMB \) are right angles. Reason: definition of perpendicular lines. 4. \( \angle CMA \cong \angle CMB \). Reason: right angle congruence theorem. 5. \( \overline{CM} \cong \overline{CM} \). Reason: reflexive property of congruence. 6. \( \triangle CMA \cong \triangle CMB \). Reason: SAS. 7. \( \overline{CA} \cong \overline{CB} \). Reason: CPCTC. This proves the perpendicular bisector theorem for one point, which lesson 6.1 generalizes. Proved by SAS and CPCTC

7. Given \( \overline{AB} \parallel \overline{DC} \) and \( \overline{AB} \cong \overline{DC} \), prove \( \overline{AD} \parallel \overline{BC} \).
   Show the full solution

   Draw the diagonal \( \overline{AC} \), creating two triangles. 1. \( \overline{AB} \parallel \overline{DC} \); \( \overline{AB} \cong \overline{DC} \). Reason: given. 2. \( \angle BAC \cong \angle DCA \). Reason: alternate interior angles theorem, with \( \overline{AC} \) as transversal. 3. \( \overline{AC} \cong \overline{AC} \). Reason: reflexive property of congruence. 4. \( \triangle BAC \cong \triangle DCA \). Reason: SAS. 5. \( \angle BCA \cong \angle DAC \). Reason: CPCTC. 6. \( \overline{AD} \parallel \overline{BC} \). Reason: converse of the alternate interior angles theorem. Note the two directions in one proof: line 2 uses the forward theorem and line 6 uses the converse. The reason column keeps them apart. Proved by SAS, CPCTC and the converse

8. A proof reads: "1. Given. 2. CPCTC. 3. SAS." Diagnose it.
   Show the full solution

   The order is impossible. CPCTC concludes something about corresponding parts, which requires the triangles to be already known congruent. At line 2 nothing has established that, so CPCTC is not available. Worse, the proof then uses SAS at line 3, presumably drawing on the part CPCTC supplied. That makes the argument circular: the congruence is used to derive a part, and the part is used to derive the congruence. The correct order is always the same: gather three pairs from givens, definitions and properties; cite a criterion to establish congruence; then cite CPCTC to extract whatever part the question wanted. Reading the reason column alone, as lesson 2.5 recommends, catches this immediately, because CPCTC appearing before any criterion is a visible ordering error. CPCTC appears before any congruence criterion, so it cites a result that is not yet available and the argument is circular

9. Given \( \overline{AC} \) bisects \( \angle BAD \) and \( \overline{AB} \cong \overline{AD} \), prove \( C \) is equidistant from \( B \) and \( D \).
   Show the full solution

   Equidistant means \( CB = CD \), so the target is a pair of congruent segments. 1. \( \overline{AC} \) bisects \( \angle BAD \); \( \overline{AB} \cong \overline{AD} \). Reason: given. 2. \( \angle BAC \cong \angle DAC \). Reason: definition of an angle bisector. 3. \( \overline{AC} \cong \overline{AC} \). Reason: reflexive property of congruence. 4. \( \triangle BAC \cong \triangle DAC \). Reason: SAS, with \( \angle A \) included between \( \overline{AB} \) and \( \overline{AC} \). 5. \( \overline{CB} \cong \overline{CD} \). Reason: CPCTC. 6. \( CB = CD \). Reason: definition of congruent segments. 7. \( C \) is equidistant from \( B \) and \( D \). Reason: definition of equidistant. The last two lines are the language conversion the method mentions: CPCTC gave congruent segments, and the question asked about distance, so the definitions carry it across. Proved by SAS and CPCTC

10. Given \( \overline{AD} \cong \overline{BC} \) and \( \angle DAB \cong \angle CBA \), prove \( \overline{AC} \cong \overline{BD} \), and explain which two triangles to use and why the obvious pair fails.
    Show the full solution

    Choosing the triangles. The goal involves \( \overline{AC} \) and \( \overline{BD} \), the two diagonals of quadrilateral \( ABCD \). The triangles containing them as sides are \( \triangle DAB \) and \( \triangle CBA \), which overlap and share the side \( \overline{AB} \). The obvious pair, \( \triangle ABC \) and \( \triangle ABD \), is the same two triangles named differently; what matters is getting the correspondence right so that \( \overline{AC} \) and \( \overline{BD} \) come out as corresponding parts. 1. \( \overline{AD} \cong \overline{BC} \); \( \angle DAB \cong \angle CBA \). Reason: given. 2. \( \overline{AB} \cong \overline{BA} \). Reason: reflexive property of congruence. 3. \( \triangle DAB \cong \triangle CBA \). Reason: SAS. 4. \( \overline{DB} \cong \overline{CA} \). Reason: CPCTC. Checking the included angle at line 3. In \( \triangle DAB \) the sides in hand are \( \overline{AD} \) and \( \overline{AB} \), meeting at \( A \), so \( \angle DAB \) is included. In \( \triangle CBA \) the sides are \( \overline{BC} \) and \( \overline{BA} \), meeting at \( B \), so \( \angle CBA \) is included. Legitimate. Checking the correspondence at line 4. The statement \( \triangle DAB \cong \triangle CBA \) matches \( D \to C \), \( A \to B \), \( B \to A \). So \( \overline{DB} \), joining the first and third vertices, corresponds to \( \overline{CA} \), joining the first and third of the other. Correct, and that is exactly the pair the question asked about. Had the correspondence been written as \( \triangle DAB \cong \triangle CAB \), the letters would claim \( A \to A \), which the shared-side argument does not support, and line 4 would have produced the wrong pair. This figure is an isosceles trapezoid when \( \overline{AB} \parallel \overline{DC} \), and the result proved here is the theorem that its diagonals are congruent, which unit 7 cites. Use \( \triangle DAB \) and \( \triangle CBA \) with SAS, then CPCTC; the correspondence must send \( A \to B \) for the diagonals to correspond

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What property builds a congruent pair by adding a shared piece?
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   The addition property of equality, with the segment or angle addition postulate

2. If \( AB = CD \), what can you add to both to relate \( AC \) and \( BD \)?
   Show the full solution

   \( BC \), the shared piece

3. What is the first step when a figure is confusing?
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   Redraw the two triangles separately

4. Two overlapping segments look equal. Is that a reason?
   Show the full solution

   No

5. If \( m\angle ABD = m\angle CBE \) and they share \( \angle CBD \), what follows about \( \angle ABC \) and \( \angle DBE \)?
   Show the full solution

   Subtract the shared angle from each. They are congruent

6. \( A \), \( B \), \( C \), \( D \) are collinear in order with \( \overline{AC} \cong \overline{BD} \). Prove \( \overline{AB} \cong \overline{CD} \).
   Show the full solution

   This is the subtraction version of the worked example. 1. \( \overline{AC} \cong \overline{BD} \). Reason: given. 2. \( AC = BD \). Reason: definition of congruent segments. 3. \( AC = AB + BC \) and \( BD = BC + CD \). Reason: segment addition postulate. 4. \( AB + BC = BC + CD \). Reason: substitution property of equality. 5. \( AB = CD \). Reason: subtraction property of equality, subtracting \( BC \). 6. \( \overline{AB} \cong \overline{CD} \). Reason: definition of congruent segments. Proved by subtraction

7. \( \overrightarrow{BD} \) and \( \overrightarrow{BE} \) lie in the interior of \( \angle ABC \) with \( \angle ABD \cong \angle CBE \). Prove \( \angle ABE \cong \angle CBD \).
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   The two target angles overlap in the shared angle \( \angle DBE \). 1. \( \angle ABD \cong \angle CBE \). Reason: given. 2. \( m\angle ABD = m\angle CBE \). Reason: definition of congruent angles. 3. \( m\angle DBE = m\angle DBE \). Reason: reflexive property of equality. 4. \( m\angle ABD + m\angle DBE = m\angle CBE + m\angle DBE \). Reason: addition property of equality. 5. \( m\angle ABE = m\angle ABD + m\angle DBE \) and \( m\angle CBD = m\angle CBE + m\angle EBD \). Reason: angle addition postulate. 6. \( m\angle ABE = m\angle CBD \). Reason: substitution property of equality. 7. \( \angle ABE \cong \angle CBD \). Reason: definition of congruent angles. Proved by the angle addition version of the same technique

8. Explain why overlapping figures are harder than separated ones.
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   Three difficulties compound. First, the shared parts are invisible. A side belonging to both triangles looks like one line in the drawing, so a student scanning for three pairs may not notice it is available twice. The reflexive property line is the fix, and it is omitted more often here than anywhere else. Second, the parts that correspond are not adjacent on the page. In two separated triangles the corresponding sides are in matching positions and easy to pair. In an overlapping figure a side of one triangle may run through the middle of the other, so the correspondence has to be worked out from the vertex letters rather than seen. Third, some congruent pairs have to be built rather than read off. Two overlapping segments are congruent only because equal pieces were added to a shared piece, and that requires an addition or subtraction argument of several lines before a criterion can be cited at all. The remedy for all three is the same: redraw the triangles separately, list the vertices of each, and identify what is shared before looking for pairs. Shared parts are easy to miss, corresponding parts are not in matching positions, and some pairs must be built by addition rather than read from the diagram

9. Given \( \overline{AB} \cong \overline{AC} \) and \( \overline{AD} \cong \overline{AE} \) with \( D \) on \( \overline{AB} \) and \( E \) on \( \overline{AC} \), prove \( \overline{DB} \cong \overline{EC} \).
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   1. \( \overline{AB} \cong \overline{AC} \); \( \overline{AD} \cong \overline{AE} \); \( D \) on \( \overline{AB} \); \( E \) on \( \overline{AC} \). Reason: given. 2. \( AB = AC \) and \( AD = AE \). Reason: definition of congruent segments. 3. \( AD + DB = AB \) and \( AE + EC = AC \). Reason: segment addition postulate. 4. \( AD + DB = AE + EC \). Reason: substitution property of equality, using \( AB = AC \). 5. \( AE + DB = AE + EC \). Reason: substitution property of equality, using \( AD = AE \). 6. \( DB = EC \). Reason: subtraction property of equality. 7. \( \overline{DB} \cong \overline{EC} \). Reason: definition of congruent segments. Proved by subtracting the congruent pieces from the congruent wholes

10. In a figure, \( \triangle ABE \) and \( \triangle ACD \) overlap with \( \overline{AB} \cong \overline{AC} \), \( \angle A \) shared, and \( \overline{AE} \cong \overline{AD} \). Prove \( \overline{BE} \cong \overline{CD} \) and then that \( \triangle BDF \cong \triangle CEF \), where \( F \) is the intersection of \( \overline{BE} \) and \( \overline{CD} \)? Determine what extra information is needed.
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    First part, which the given supports. 1. \( \overline{AB} \cong \overline{AC} \); \( \overline{AE} \cong \overline{AD} \). Reason: given. 2. \( \angle A \cong \angle A \). Reason: reflexive property of congruence. 3. \( \triangle ABE \cong \triangle ACD \). Reason: SAS, with \( \angle A \) included between \( \overline{AB} \) and \( \overline{AE} \) in the first triangle and between \( \overline{AC} \) and \( \overline{AD} \) in the second. 4. \( \overline{BE} \cong \overline{CD} \). Reason: CPCTC. Second part, which needs more work. The triangles \( \triangle BDF \) and \( \triangle CEF \) sit at the intersection point. From line 3 we also get \( \angle ABE \cong \angle ACD \) and \( \angle AEB \cong \angle ADC \) by CPCTC, and the vertical angles at \( F \) give \( \angle BFD \cong \angle CFE \). That is two angle pairs, so AAS or ASA would finish if one side pair were available. The natural candidate is \( \overline{BD} \cong \overline{CE} \), which follows by the subtraction argument of the previous problem: \( AB = AC \) and \( AD = AE \), with \( D \) on \( \overline{AB} \) and \( E \) on \( \overline{AC} \), give \( DB = EC \). What is needed. The problem as stated does not say that \( D \) lies on \( \overline{AB} \) and \( E \) on \( \overline{AC} \). Without that, the subtraction argument is unavailable and no side pair can be built, so the second congruence cannot be proved. With it, the proof completes by AAS. This is worth noticing as a habit: when a proof stalls, the question is usually which given is missing rather than which theorem was forgotten. The first congruence follows by SAS and CPCTC; the second needs the additional given that \( D \) lies on \( \overline{AB} \) and \( E \) on \( \overline{AC} \), and then follows by AAS

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. In \( \triangle ABC \) with \( \overline{AB} \cong \overline{AC} \), which angles are congruent?
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   \( \angle B \) and \( \angle C \)

2. An isosceles triangle has a vertex angle of \( 80^\circ \). Find each base angle.
   Show the full solution

   \( (180 - 80) \div 2 \). \( 50^\circ \)

3. What is each angle of an equilateral triangle?
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   \( 60^\circ \)

4. Two angles of a triangle measure \( 45^\circ \) each. What follows about the sides?
   Show the full solution

   By the converse of the base angles theorem. The sides opposite them are congruent

5. An isosceles triangle has base angles of \( 72^\circ \). Find the vertex angle.
   Show the full solution

   \( 180 - 144 \). \( 36^\circ \)

6. In \( \triangle ABC \), \( \overline{AB} \cong \overline{AC} \), \( m\angle B = (3x + 10)^\circ \) and \( m\angle C = (5x - 20)^\circ \). Find all three angles.
   Show the full solution

   The congruent sides are \( \overline{AB} \) and \( \overline{AC} \), so the base angles are \( \angle B \) and \( \angle C \), congruent by the base angles theorem. \( 3x + 10 = 5x - 20 \), giving \( 30 = 2x \) and \( x = 15 \). Each base angle measures \( 3(15) + 10 = 55^\circ \), confirmed by \( 5(15) - 20 = 55^\circ \). Vertex angle: \( m\angle A = 180 - 55 - 55 = 70^\circ \). Check: \( 70 + 55 + 55 = 180 \). Correct. \( 70^\circ \), \( 55^\circ \), \( 55^\circ \)

7. An isosceles triangle has one angle of \( 100^\circ \). Find the other two, and explain why only one arrangement is possible.
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   The \( 100^\circ \) angle must be the vertex angle. If it were a base angle, the other base angle would also measure \( 100^\circ \) and the two alone would sum to \( 200^\circ \), exceeding the total of \( 180^\circ \) available. Impossible. So \( 100^\circ \) is the vertex angle, and the two base angles share the remaining \( 80^\circ \) equally: \( 40^\circ \) each. Check: \( 100 + 40 + 40 = 180 \). Correct. The general principle: an obtuse or right angle in an isosceles triangle must be the vertex angle, since two of them would already exhaust or exceed the angle sum. Only an acute angle can be ambiguous. \( 40^\circ \) and \( 40^\circ \); an obtuse angle cannot be a base angle

8. An isosceles triangle has one angle of \( 50^\circ \). Find all the possibilities.
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   Since \( 50^\circ \) is acute, both arrangements are possible and both must be given. Case one: \( 50^\circ \) is the vertex angle. The two base angles share \( 180 - 50 = 130^\circ \), so each measures \( 65^\circ \). The triangle is \( 50^\circ \), \( 65^\circ \), \( 65^\circ \). Case two: \( 50^\circ \) is a base angle. Then the other base angle is also \( 50^\circ \), and the vertex angle is \( 180 - 100 = 80^\circ \). The triangle is \( 50^\circ \), \( 50^\circ \), \( 80^\circ \). Both check: each set sums to \( 180^\circ \) and each contains two equal angles. Giving only one case is the error here. The question did not say which angle was named, and both are genuinely possible triangles. Two possibilities: \( 50, 65, 65 \) or \( 50, 50, 80 \)

9. Explain why the base angles theorem and its converse can both be true when most theorems have false converses.
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   Nothing guarantees a converse is true; it has to be proved separately, and often it cannot be. The reason both hold here is that the isosceles condition and the equal-angle condition are genuinely equivalent for triangles, and each direction has its own proof. The forward proof, in the worked example, bisects the vertex angle and uses SAS to get two congruent triangles, then CPCTC for the angles. The converse proof runs similarly but uses AAS. Given \( \angle B \cong \angle C \) in \( \triangle ABC \), draw the bisector of \( \angle A \) meeting \( \overline{BC} \) at \( D \). Then \( \angle B \cong \angle C \) is given, \( \angle BAD \cong \angle CAD \) by the bisector, and \( \overline{AD} \) is shared. That is AAS, giving \( \triangle BAD \cong \triangle CAD \) and then \( \overline{AB} \cong \overline{AC} \) by CPCTC. So both directions are provable, by different criteria, and the statement can be written as a biconditional: a triangle has two congruent sides if and only if it has two congruent angles. Contrast a theorem whose converse fails, such as "a linear pair is supplementary." No proof of the converse exists because the converse is false, and a counterexample settles it. The practical point remains the one from lesson 2.2: a converse's truth is never inherited and must always be checked. Each direction has its own proof, the forward one by SAS and the converse by AAS, so the conditions are genuinely equivalent

10. In \( \triangle ABC \), \( \overline{AB} \cong \overline{AC} \) and \( D \) is the midpoint of \( \overline{BC} \). Prove that \( \overline{AD} \) is both an altitude and the bisector of \( \angle A \).
    Show the full solution

    1. \( \overline{AB} \cong \overline{AC} \); \( D \) is the midpoint of \( \overline{BC} \). Reason: given. 2. \( \overline{BD} \cong \overline{DC} \). Reason: definition of midpoint. 3. \( \overline{AD} \cong \overline{AD} \). Reason: reflexive property of congruence. 4. \( \triangle ABD \cong \triangle ACD \). Reason: SSS. 5. \( \angle BAD \cong \angle CAD \). Reason: CPCTC. 6. \( \overline{AD} \) bisects \( \angle BAC \). Reason: definition of an angle bisector. 7. \( \angle ADB \cong \angle ADC \). Reason: CPCTC. 8. \( \angle ADB \) and \( \angle ADC \) form a linear pair. Reason: \( D \) lies on \( \overline{BC} \), so the two non-shared sides form a line. 9. \( \angle ADB \) and \( \angle ADC \) are supplementary. Reason: linear pair theorem. 10. \( m\angle ADB = 90 \). Reason: two congruent supplementary angles are right angles, from the theorem at the end of lesson 1.4. 11. \( \overline{AD} \perp \overline{BC} \). Reason: definition of perpendicular lines. 12. \( \overline{AD} \) is an altitude. Reason: definition of an altitude, a segment from a vertex perpendicular to the opposite side. This completes the picture from lesson 5.4: in an isosceles triangle, the median to the base, the altitude to the base, the angle bisector from the vertex, and the perpendicular bisector of the base are all the same segment. Any one of those four conditions implies the other three, which is why isosceles triangles appear so often in proof problems. Proved: SSS gives the congruence, and CPCTC yields both the angle bisector and the right angle

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A point is on the perpendicular bisector of \( \overline{AB} \). What follows?
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   It is equidistant from \( A \) and \( B \)

2. What is the circumcenter equidistant from?
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   The three vertices

3. Where is the circumcenter of a right triangle?
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   At the midpoint of the hypotenuse

4. How many perpendicular bisectors must you intersect to find the circumcenter?
   Show the full solution

   Two

5. \( P \) is equidistant from \( A \) and \( B \). What follows?
   Show the full solution

   By the converse. \( P \) is on the perpendicular bisector of \( \overline{AB} \)

6. Find the circumcenter of the triangle with vertices \( (0,0) \), \( (8,0) \), \( (0,6) \).
   Show the full solution

   Perpendicular bisector of the horizontal side from \( (0,0) \) to \( (8,0) \): the vertical line \( x = 4 \). Perpendicular bisector of the vertical side from \( (0,0) \) to \( (0,6) \): the horizontal line \( y = 3 \). They meet at \( (4, 3) \). Check: distance to \( (0,0) \) is \( \sqrt{16 + 9} = 5 \); to \( (8,0) \) is \( \sqrt{16 + 9} = 5 \); to \( (0,6) \) is \( \sqrt{16 + 9} = 5 \). All equal. The triangle is right at the origin, and \( (4,3) \) is the midpoint of the hypotenuse from \( (8,0) \) to \( (0,6) \), as the rule predicts. \( (4, 3) \), with circumradius 5

7. A triangle has a circumradius of 13 and is right. Find the hypotenuse.
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   In a right triangle the circumcenter is the midpoint of the hypotenuse, so the circumradius is half the hypotenuse. Therefore the hypotenuse is \( 2 \times 13 = 26 \). The reason: the midpoint of the hypotenuse is by definition equidistant from the two endpoints of the hypotenuse, at half its length each, and the circumcenter property says it is the same distance from the third vertex. 26

8. Prove the converse of the perpendicular bisector theorem.
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   Given: \( PA = PB \). Prove: \( P \) lies on the perpendicular bisector of \( \overline{AB} \). 1. \( PA = PB \). Reason: given. 2. Draw \( \overline{PM} \perp \overline{AB} \) with \( M \) on \( \overline{AB} \). Reason: through any point there is a perpendicular to a given line. 3. \( \angle PMA \) and \( \angle PMB \) are right angles. Reason: definition of perpendicular lines. 4. \( \triangle PMA \) and \( \triangle PMB \) are right triangles. Reason: definition of a right triangle. 5. \( \overline{PA} \cong \overline{PB} \). Reason: definition of congruent segments, from line 1. 6. \( \overline{PM} \cong \overline{PM} \). Reason: reflexive property of congruence. 7. \( \triangle PMA \cong \triangle PMB \). Reason: HL, with \( \overline{PA} \) and \( \overline{PB} \) the hypotenuses. 8. \( \overline{AM} \cong \overline{MB} \). Reason: CPCTC. 9. \( M \) is the midpoint of \( \overline{AB} \). Reason: definition of midpoint. 10. \( \overline{PM} \) is the perpendicular bisector of \( \overline{AB} \), so \( P \) lies on it. Reason: definition of a perpendicular bisector, using lines 2 and 9. Note that the converse needed HL while the forward direction needed SAS. Different criteria for different directions is typical. Proved by HL and CPCTC

9. Explain why the three perpendicular bisectors meet at one point.
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   Take \( \triangle ABC \) and let \( P \) be the intersection of the perpendicular bisectors of \( \overline{AB} \) and \( \overline{BC} \). Those two lines are not parallel, since the sides they bisect are not parallel, so they do meet. Because \( P \) is on the perpendicular bisector of \( \overline{AB} \), the theorem gives \( PA = PB \). Because \( P \) is on the perpendicular bisector of \( \overline{BC} \), the theorem gives \( PB = PC \). By the transitive property, \( PA = PC \). So \( P \) is equidistant from \( A \) and \( C \). By the converse of the theorem, \( P \) must lie on the perpendicular bisector of \( \overline{AC} \). So the third bisector passes through \( P \) as well, and all three are concurrent. Both directions of the theorem were needed: the forward direction to get the distances, and the converse to conclude the third line contains \( P \). That is why proving the converse in the previous problem was worth the trouble. The intersection of two bisectors is equidistant from all three vertices, and the converse then forces the third bisector through it

10. Find the circumcenter of the triangle with vertices \( (1,1) \), \( (5,1) \), \( (3,5) \), and determine whether the triangle is acute, right or obtuse.
    Show the full solution

    First bisector. The side from \( (1,1) \) to \( (5,1) \) is horizontal with midpoint \( (3, 1) \), so its perpendicular bisector is \( x = 3 \). Second bisector. The side from \( (1,1) \) to \( (3,5) \) has midpoint \( (2, 3) \) and slope \( \dfrac{5-1}{3-1} = 2 \), so the perpendicular slope is \( -\dfrac{1}{2} \) and the bisector is \( y - 3 = -\dfrac{1}{2}(x - 2) \), that is \( y = -\dfrac{1}{2}x + 4 \). Intersect. At \( x = 3 \): \( y = -1.5 + 4 = 2.5 \). The circumcenter is \( (3, 2.5) \). Verify. To \( (1,1) \): differences \( -2 \) and \( -1.5 \), so \( \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5 \). To \( (5,1) \): differences 2 and \( -1.5 \), so \( \sqrt{6.25} = 2.5 \). To \( (3,5) \): differences 0 and 2.5, so 2.5. All equal. Classify the triangle. The circumcenter \( (3, 2.5) \) lies inside the triangle, whose vertices span \( y \) from 1 to 5 and whose base runs from \( x = 1 \) to \( x = 5 \) at \( y = 1 \). A circumcenter inside means the triangle is acute. Confirming by side lengths: the base is 4; the two equal sides are \( \sqrt{4 + 16} = \sqrt{20} \approx 4.47 \). The largest side is \( \sqrt{20} \), and \( 4^2 + 20 = 36 \gt 20 \), so by the converse of the Pythagorean theorem in lesson 9.1 the largest angle is acute. Consistent. Circumcenter \( (3, 2.5) \), circumradius 2.5; the triangle is acute

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is the incenter equidistant from?
   Show the full solution

   The three sides

2. Is the incenter always inside the triangle?
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   Yes

3. What three lines meet at the incenter?
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   The three angle bisectors

4. What circle is centered at the incenter?
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   The inscribed circle

5. What does "distance to a side" mean?
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   The perpendicular distance

6. A triangle has area 30 and perimeter 30. Find its inradius.
   Show the full solution

   Half the perimeter is \( s = 15 \). \( r = \dfrac{\text{Area}}{s} = \dfrac{30}{15} = 2 \). 2

7. A right triangle has legs 6 and 8. Find its inradius two ways.
   Show the full solution

   The hypotenuse is \( \sqrt{36 + 64} = 10 \). First way. Area is \( \dfrac{1}{2}(6)(8) = 24 \). Perimeter is \( 6 + 8 + 10 = 24 \), so \( s = 12 \). Then \( r = \dfrac{24}{12} = 2 \). Second way. For a right triangle, \( r = \dfrac{a + b - c}{2} = \dfrac{6 + 8 - 10}{2} = \dfrac{4}{2} = 2 \). The two agree. Note this triangle is the 3-4-5 of the worked example doubled, and its inradius doubled too, from 1 to 2, which is what scaling by a factor of 2 does to every length. 2

8. Explain the difference between the circumcenter and the incenter in one paragraph.
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   They are built from different lines and have different equidistance properties. The circumcenter is where the three perpendicular bisectors of the sides meet. It is equidistant from the three vertices, so it is the center of the circle passing through them, and it may lie inside, on, or outside the triangle depending on whether the triangle is acute, right or obtuse. The incenter is where the three angle bisectors meet. It is equidistant from the three sides, so it is the center of the circle touching them, and it always lies inside, because every angle bisector runs through the interior. The two coincide only for an equilateral triangle, where the symmetry makes every special point the same. The reading test for a question: if it says vertices or "circle through the vertices," the answer is the circumcenter; if it says sides or "circle inside," the answer is the incenter. Circumcenter: perpendicular bisectors, equidistant from vertices, may be outside. Incenter: angle bisectors, equidistant from sides, always inside

9. An equilateral triangle has side 6. Find its inradius and circumradius, and comment.
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   Area. An equilateral triangle of side \( a \) has height \( \dfrac{a\sqrt{3}}{2} \), so here the height is \( 3\sqrt{3} \) and the area is \( \dfrac{1}{2}(6)(3\sqrt{3}) = 9\sqrt{3} \approx 15.59 \). Inradius. The perimeter is 18, so \( s = 9 \) and \( r = \dfrac{9\sqrt{3}}{9} = \sqrt{3} \approx 1.73 \). Circumradius. In an equilateral triangle the center is also the centroid, which lies two thirds of the way along each median from the vertex. The median is the height, \( 3\sqrt{3} \), so the circumradius is \( \dfrac{2}{3}(3\sqrt{3}) = 2\sqrt{3} \approx 3.46 \). Comment. The circumradius is exactly twice the inradius, and the two circles are concentric because the incenter and circumcenter coincide. That happens only for equilateral triangles, where every median, altitude, angle bisector and perpendicular bisector is the same line, so all four centers of this unit fall together. Check: the inradius plus the circumradius is \( 3\sqrt{3} \), the full height, which is what the geometry requires since the center lies on the altitude at distance \( r \) from the base and \( R \) from the opposite vertex. Inradius \( \sqrt{3} \approx 1.73 \), circumradius \( 2\sqrt{3} \approx 3.46 \), concentric because all centers coincide

10. A point inside an angle is 5 units from one side and 5 units from the other. What can be concluded, and what extra information would be needed to locate it exactly?
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    What follows. By the converse of the angle bisector theorem, a point in the interior of an angle equidistant from both sides lies on the bisector of that angle. So the point is somewhere on the bisector. Why that is not enough. The bisector is a ray containing infinitely many points, and every one of them is equidistant from the two sides, at distances increasing with the distance from the vertex. Knowing the common distance is 5 narrows it to one point on the ray, provided the angle is known, because the distance from the vertex is then determined by trigonometry: if the angle measures \( 2\theta \), the point sits at distance \( \dfrac{5}{\sin\theta} \) from the vertex. What would pin it down. Either the measure of the angle, which fixes the position along the bisector as just described, or the distance from the vertex directly, or a third distance to some other line in the figure. This is the same situation as the circumcenter: one condition gives a line of candidates, and a second condition intersects it down to a point. The incenter is located by two such conditions, one from each of two angle bisectors. It lies on the angle bisector; its exact position needs the angle measure or the distance from the vertex

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does a median join?
   Show the full solution

   A vertex to the midpoint of the opposite side

2. What is the intersection of the medians called?
   Show the full solution

   The centroid

3. Find the centroid of the triangle with vertices \( (0,0) \), \( (3,0) \), \( (0,6) \).
   Show the full solution

   Average each coordinate. \( (1, 2) \)

4. A median is 18 long. Find the vertex-to-centroid piece.
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   Two thirds of 18. 12

5. Is the centroid always inside the triangle?
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   Yes

6. Find the centroid of the triangle with vertices \( (2, 5) \), \( (8, 1) \), \( (-1, 3) \).
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   \( x \): \( \dfrac{2 + 8 + (-1)}{3} = \dfrac{9}{3} = 3 \). \( y \): \( \dfrac{5 + 1 + 3}{3} = \dfrac{9}{3} = 3 \). \( (3, 3) \)

7. The centroid-to-midpoint piece of a median is 7. Find the whole median.
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   That short piece is one third of the median, since the pieces are in a two-to-one ratio and therefore split the median into three equal parts. Whole median: \( 3 \times 7 = 21 \). The vertex-to-centroid piece is \( 2 \times 7 = 14 \), and \( 14 + 7 = 21 \). Consistent. 21

8. Two vertices of a triangle are \( (1, 2) \) and \( (7, 4) \), and the centroid is \( (4, 5) \). Find the third vertex.
   Show the full solution

   The centroid formula averages the three vertices, so multiply back. For \( x \): \( \dfrac{1 + 7 + x}{3} = 4 \), so \( 8 + x = 12 \) and \( x = 4 \). For \( y \): \( \dfrac{2 + 4 + y}{3} = 5 \), so \( 6 + y = 15 \) and \( y = 9 \). The third vertex is \( (4, 9) \). Check: \( \dfrac{1 + 7 + 4}{3} = 4 \) and \( \dfrac{2 + 4 + 9}{3} = 5 \). Correct. \( (4, 9) \)

9. Explain why the centroid formula is the average of the vertices.
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   Take vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), \( C(x_3, y_3) \). The midpoint of \( \overline{BC} \) is \( M = \left( \dfrac{x_2 + x_3}{2}, \dfrac{y_2 + y_3}{2} \right) \). The centroid is two thirds of the way from \( A \) to \( M \), so its coordinates are \( A + \dfrac{2}{3}(M - A) \). Working on the \( x \) coordinate: \( x_1 + \dfrac{2}{3}\left( \dfrac{x_2 + x_3}{2} - x_1 \right) = x_1 + \dfrac{x_2 + x_3}{3} - \dfrac{2x_1}{3} = \dfrac{x_1}{3} + \dfrac{x_2 + x_3}{3} = \dfrac{x_1 + x_2 + x_3}{3} \). The same computation on \( y \) gives \( \dfrac{y_1 + y_2 + y_3}{3} \). So the two-to-one ratio is exactly what produces the average, and the symmetry of the result explains the concurrency: the expression treats the three vertices identically, so starting from \( B \) or \( C \) instead gives the same point. That symmetry is a genuine proof of concurrency for the medians, and it is shorter than the synthetic argument. Two thirds of the way from a vertex to the opposite midpoint works out algebraically to the average, and the symmetric result proves the three medians concur

10. In \( \triangle ABC \), the median from \( A \) meets \( \overline{BC} \) at \( M \), and \( G \) is the centroid with \( AG = 4x - 2 \) and \( GM = x + 3 \). Find the length of the median.
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    The vertex-to-centroid piece is twice the centroid-to-midpoint piece: \( AG = 2 \cdot GM \). Substituting: \( 4x - 2 = 2(x + 3) \), so \( 4x - 2 = 2x + 6 \), giving \( 2x = 8 \) and \( x = 4 \). Then \( AG = 4(4) - 2 = 14 \) and \( GM = 4 + 3 = 7 \). Check the ratio: \( 14 = 2 \times 7 \). Correct, and the long piece is at the vertex as required. The whole median is \( AM = AG + GM = 14 + 7 = 21 \). A second check: the median should be three times the short piece, and \( 3 \times 7 = 21 \). Consistent. Note the trap: setting \( GM = 2 \cdot AG \) instead gives \( x + 3 = 8x - 4 \), so \( x = 1 \), with \( AG = 2 \) and \( GM = 4 \). That arrangement has the long piece at the midpoint end, which contradicts the theorem, and checking which piece touches the vertex catches it. The median is 21, with \( AG = 14 \) and \( GM = 7 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is an altitude?
   Show the full solution

   A segment from a vertex perpendicular to the line containing the opposite side

2. What is the intersection of the altitudes called?
   Show the full solution

   The orthocenter

3. Where is the orthocenter of a right triangle?
   Show the full solution

   At the right-angle vertex

4. Is an altitude always inside the triangle?
   Show the full solution

   No; in an obtuse triangle two of them fall outside

5. Which two of the four centers are always inside?
   Show the full solution

   The incenter and the centroid

6. Find the altitude from \( (0, 5) \) to the side joining \( (0,0) \) and \( (4, 0) \).
   Show the full solution

   The side is horizontal with slope 0, so the altitude is vertical through \( (0, 5) \), that is the line \( x = 0 \). Its foot is where \( x = 0 \) meets \( y = 0 \), the point \( (0, 0) \). So the altitude runs from \( (0,5) \) to \( (0,0) \) and has length 5. Notice its foot coincides with a vertex, which happens exactly when the triangle has a right angle there, and indeed the sides from \( (0,0) \) are vertical and horizontal. The segment from \( (0,5) \) to \( (0,0) \), of length 5

7. In a right triangle with legs along the axes, name the three altitudes.
   Show the full solution

   Place the right angle at the origin with vertices \( (0,0) \), \( (a, 0) \) and \( (0, b) \). The altitude from \( (0,b) \) is perpendicular to the horizontal side, so it is the vertical segment from \( (0,b) \) to \( (0,0) \), which is the leg itself. The altitude from \( (a,0) \) is perpendicular to the vertical side, so it is the horizontal segment from \( (a,0) \) to \( (0,0) \), the other leg. The altitude from \( (0,0) \) is perpendicular to the hypotenuse and runs into the interior to meet it. So two of the three altitudes are the legs, and all three meet at \( (0,0) \), the right-angle vertex. That is why the orthocenter of a right triangle sits there. The two legs, plus the perpendicular from the right angle to the hypotenuse

8. Find the orthocenter of the triangle with vertices \( (0,0) \), \( (6,0) \), \( (2, 4) \).
   Show the full solution

   Altitude from \( (2,4) \). The opposite side is horizontal, so this altitude is vertical: \( x = 2 \). Altitude from \( (0,0) \). The opposite side joins \( (6,0) \) and \( (2,4) \), with slope \( \dfrac{4 - 0}{2 - 6} = -1 \). The perpendicular slope is 1, so the altitude is \( y = x \). Intersect. At \( x = 2 \), \( y = 2 \). The orthocenter is \( (2, 2) \). Verify with the third. The side from \( (0,0) \) to \( (2,4) \) has slope 2, so the altitude from \( (6,0) \) has slope \( -\dfrac{1}{2} \): \( y = -\dfrac{1}{2}(x - 6) \). At \( x = 2 \): \( y = -\dfrac{1}{2}(-4) = 2 \). Passes through \( (2,2) \). Concurrent. \( (2, 2) \)

9. Explain why an altitude and a median from the same vertex coincide only in special cases.
   Show the full solution

   They are defined by different conditions. The median is determined by where it lands, at the midpoint of the opposite side. The altitude is determined by the angle at which it meets, perpendicular to that side. There is no reason these should pick out the same segment. They coincide exactly when the perpendicular from the vertex happens to hit the midpoint, which by the converse of the perpendicular bisector theorem means the vertex is equidistant from the two endpoints of that side. In other words, the triangle is isosceles with that side as the base. Lesson 5.7 proved the full version: in an isosceles triangle, the altitude, median, angle bisector from the vertex angle, and perpendicular bisector of the base are all the same segment. In an equilateral triangle this holds for all three vertices, which is why all four centers of this unit coincide there. In a scalene triangle they never coincide, as the worked example showed with a foot at \( (2,0) \) and a midpoint at \( (4,0) \). They coincide only when the triangle is isosceles with that side as the base, since that is when the perpendicular from the vertex reaches the midpoint

10. An obtuse triangle has vertices \( (0,0) \), \( (6,0) \), \( (8, 3) \). Find the orthocenter and confirm it lies outside.
    Show the full solution

    Confirm the triangle is obtuse first. Side lengths: from \( (0,0) \) to \( (6,0) \) is 6; from \( (6,0) \) to \( (8,3) \) is \( \sqrt{4 + 9} = \sqrt{13} \approx 3.61 \); from \( (0,0) \) to \( (8,3) \) is \( \sqrt{64 + 9} = \sqrt{73} \approx 8.54 \). The longest side is \( \sqrt{73} \), opposite the vertex \( (6,0) \). Testing: \( 6^2 + 13 = 49 \), which is less than 73, so by the converse of the Pythagorean theorem that angle is obtuse. Altitude from \( (8,3) \). The opposite side is horizontal, so the altitude is vertical: \( x = 8 \). Note its foot is at \( (8, 0) \), which is outside the segment from \( (0,0) \) to \( (6,0) \). That is the "line containing" clause of the definition doing its work. Altitude from \( (0,0) \). The opposite side joins \( (6,0) \) and \( (8,3) \), with slope \( \dfrac{3}{2} \). The perpendicular slope is \( -\dfrac{2}{3} \), so the altitude is \( y = -\dfrac{2}{3}x \). Intersect. At \( x = 8 \): \( y = -\dfrac{16}{3} \approx -5.33 \). The orthocenter is \( \left( 8, -\dfrac{16}{3} \right) \). Confirm it is outside. The triangle lies entirely in the region with \( y \) between 0 and 3, and \( x \) between 0 and 8. The orthocenter has \( y \approx -5.33 \), well below the triangle. Outside, as the rule for obtuse triangles predicts. \( \left( 8, -\frac{16}{3} \right) \), outside the triangle

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does a midsegment join?
   Show the full solution

   The midpoints of two sides of a triangle

2. A midsegment is 7. How long is the third side?
   Show the full solution

   14

3. A side is 22. How long is the midsegment parallel to it?
   Show the full solution

   11

4. How many midsegments does a triangle have?
   Show the full solution

   Three

5. What is the relationship between a midsegment and the third side, besides length?
   Show the full solution

   They are parallel

6. Find the midsegment of the triangle \( (0,0) \), \( (10,0) \), \( (4,8) \) parallel to the horizontal side, and verify the theorem.
   Show the full solution

   The midsegment parallel to the side from \( (0,0) \) to \( (10,0) \) joins the midpoints of the other two sides. Midpoint of the side from \( (0,0) \) to \( (4,8) \): \( (2, 4) \). Midpoint of the side from \( (10,0) \) to \( (4,8) \): \( (7, 4) \). The midsegment runs from \( (2,4) \) to \( (7,4) \), horizontal with length 5. The third side is horizontal with length 10. Parallel: both slopes are 0. Half: \( 5 = \dfrac{1}{2}(10) \). Both verified. From \( (2,4) \) to \( (7,4) \), length 5, parallel to and half of the side of length 10

7. A triangle has perimeter 36. Find the perimeter of the triangle formed by its three midsegments.
   Show the full solution

   Each midsegment is half the side it is parallel to, so the three midsegments are halves of the three sides. Their sum is therefore half the sum of the sides: \( \dfrac{1}{2}(36) = 18 \). 18

8. In \( \triangle ABC \), \( D \) is the midpoint of \( \overline{AB} \) and \( E \) is the midpoint of \( \overline{AC} \). If \( DE = 3x - 4 \) and \( BC = 4x + 6 \), find both lengths.
   Show the full solution

   \( \overline{DE} \) is a midsegment, so \( BC = 2 \cdot DE \): \( 4x + 6 = 2(3x - 4) \), giving \( 4x + 6 = 6x - 8 \), so \( 14 = 2x \) and \( x = 7 \). Then \( DE = 3(7) - 4 = 17 \) and \( BC = 4(7) + 6 = 34 \). Check: \( 34 = 2 \times 17 \). Correct, with the midsegment the shorter of the two as it must be. The trap is writing \( DE = 2 \cdot BC \), which gives \( x = -3.2 \) and negative lengths, an immediate signal of the error. \( DE = 17 \) and \( BC = 34 \)

9. Prove the midsegment theorem using coordinates.
   Show the full solution

   Place the triangle conveniently, which is allowed because any triangle can be positioned this way by a rigid motion and rigid motions preserve everything in the claim. Let \( A(0,0) \), \( B(2b, 0) \) and \( C(2c, 2d) \), using even coefficients so the midpoints have whole-number-looking coordinates. Midpoint of \( \overline{AC} \): \( M = \left( \dfrac{0 + 2c}{2}, \dfrac{0 + 2d}{2} \right) = (c, d) \). Midpoint of \( \overline{BC} \): \( N = \left( \dfrac{2b + 2c}{2}, \dfrac{0 + 2d}{2} \right) = (b + c, d) \). Parallel. The segment \( \overline{MN} \) has both endpoints at \( y = d \), so its slope is 0. The side \( \overline{AB} \) has both endpoints at \( y = 0 \), so its slope is also 0. Equal slopes, and the lines are distinct provided \( d \neq 0 \), which holds since \( C \) is not on \( \overline{AB} \). So \( \overline{MN} \parallel \overline{AB} \). Half the length. \( MN = (b + c) - c = b \), and \( AB = 2b - 0 = 2b \). So \( MN = \dfrac{1}{2}AB \). Both conclusions hold for arbitrary \( b \), \( c \), \( d \), so the theorem is proved in general and not merely for one triangle. The choice of coordinates was a convenience, not an assumption, which is the technique coordinate proofs rely on. Proved: the midpoints both have \( y = d \), giving parallelism, and the horizontal separation is \( b \) against the side's \( 2b \)

10. The three midsegments divide a triangle into four smaller triangles. Prove they are all congruent, and find the area of each if the original has area 40.
    Show the full solution

    Setting up. Let \( \triangle ABC \) have midpoints \( D \) on \( \overline{AB} \), \( E \) on \( \overline{BC} \), \( F \) on \( \overline{AC} \). The three midsegments \( \overline{DE} \), \( \overline{EF} \), \( \overline{DF} \) create the inner triangle \( \triangle DEF \) and three corner triangles \( \triangle ADF \), \( \triangle DBE \), \( \triangle FEC \). Lengths. By the midsegment theorem, \( DE = \dfrac{1}{2}AC = AF = FC \), \( EF = \dfrac{1}{2}AB = AD = DB \), and \( DF = \dfrac{1}{2}BC = BE = EC \). Corner against inner. In \( \triangle ADF \) the sides are \( AD \), \( AF \) and \( DF \). In \( \triangle DEF \) the sides are \( DE \), \( EF \) and \( DF \). From the list above, \( AD = EF \), \( AF = DE \), and \( DF = DF \). Three pairs of congruent sides, so \( \triangle ADF \cong \triangle EFD \) by SSS. The same argument applies to the other two corners, so all four triangles are congruent to each other. Area. Four congruent triangles partition the original, so each has one quarter of its area: \( \dfrac{40}{4} = 10 \). Consistency check. The inner triangle is similar to the original with scale factor \( \dfrac{1}{2} \), and lesson 11.6 says area scales by the square of the factor, so its area should be \( \left(\dfrac{1}{2}\right)^2 \times 40 = 10 \). It agrees, which is a genuine check since the two arguments are independent. All four are congruent by SSS, and each has area 10

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. In \( \triangle ABC \), which side is opposite \( \angle B \)?
   Show the full solution

   \( \overline{AC} \)

2. The largest angle of a triangle is opposite which side?
   Show the full solution

   The longest

3. Which is the longest side of a right triangle?
   Show the full solution

   The hypotenuse

4. An exterior angle measures \( 110^\circ \). What do you know about each remote interior angle?
   Show the full solution

   Each is less than \( 110^\circ \)

5. An exterior angle measures \( 130^\circ \) and one remote interior angle is \( 50^\circ \). Find the other.
   Show the full solution

   By the exterior angle theorem, \( 130 - 50 \). \( 80^\circ \)

6. In \( \triangle PQR \), \( m\angle P = 35^\circ \) and \( m\angle Q = 85^\circ \). Order the sides.
   Show the full solution

   Third angle: \( m\angle R = 180 - 35 - 85 = 60^\circ \). Angles in order: \( \angle P \) at \( 35^\circ \), \( \angle R \) at \( 60^\circ \), \( \angle Q \) at \( 85^\circ \). Opposite sides: \( \angle P \) faces \( \overline{QR} \), \( \angle R \) faces \( \overline{PQ} \), \( \angle Q \) faces \( \overline{PR} \). \( \overline{QR} \lt \overline{PQ} \lt \overline{PR} \)

7. In \( \triangle XYZ \), \( XY = 12 \), \( YZ = 8 \), \( XZ = 15 \). Order the angles.
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   Sides in increasing order: \( YZ = 8 \), \( XY = 12 \), \( XZ = 15 \). Opposite angles: \( \overline{YZ} \) faces \( \angle X \), \( \overline{XY} \) faces \( \angle Z \), \( \overline{XZ} \) faces \( \angle Y \). So the angles in increasing order are \( \angle X \), \( \angle Z \), \( \angle Y \). Sanity check on the largest: \( 8^2 + 12^2 = 64 + 144 = 208 \), which exceeds \( 15^2 = 225 \)? No, \( 208 \lt 225 \), so \( \angle Y \) is obtuse, consistent with it being the largest. \( \angle X \lt \angle Z \lt \angle Y \)

8. In an isosceles triangle with vertex angle \( 100^\circ \), which side is longest?
   Show the full solution

   The base angles are each \( (180 - 100) \div 2 = 40^\circ \). So the angles are \( 100^\circ \), \( 40^\circ \), \( 40^\circ \), and the largest is the vertex angle. The side opposite the vertex angle is the base. So the base is the longest side, and the two congruent legs are shorter. This is worth noting because "leg" sounds like it should be longer than "base." In an isosceles triangle with an obtuse vertex angle the base is longest; with an acute vertex angle the legs are longest; with a \( 60^\circ \) vertex angle the triangle is equilateral and all three are equal. The base, since it is opposite the largest angle

9. Explain why the exterior angle inequality follows from the exterior angle theorem.
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   The exterior angle theorem says an exterior angle equals the sum of the two remote interior angles: \( m\angle \text{ext} = m\angle 1 + m\angle 2 \). Both remote interior angles are angles of a triangle, so both have positive measure. Adding a positive number to \( m\angle 1 \) produces something larger than \( m\angle 1 \), so \( m\angle \text{ext} \gt m\angle 1 \). The same argument with the roles swapped gives \( m\angle \text{ext} \gt m\angle 2 \). That is the inequality, obtained in one line from the equality. The inequality is worth stating separately because it is what comparison arguments use, and because it is the weaker statement that survives in contexts where the equality does not. The equality depends on the angle sum of a triangle being exactly \( 180^\circ \), which in turn depends on the parallel postulate; the inequality can be proved without it. The exterior angle equals a sum including the remote interior angle plus a positive quantity, so it exceeds it

10. In \( \triangle ABC \), \( m\angle A = (2x + 10)^\circ \), \( m\angle B = (3x)^\circ \), \( m\angle C = (x + 50)^\circ \). Order the sides.
    Show the full solution

    Find \( x \). The angles sum to \( 180^\circ \): \( (2x + 10) + 3x + (x + 50) = 180 \), so \( 6x + 60 = 180 \), giving \( 6x = 120 \) and \( x = 20 \). Find the angles. \( m\angle A = 2(20) + 10 = 50^\circ \). \( m\angle B = 3(20) = 60^\circ \). \( m\angle C = 20 + 50 = 70^\circ \). Check: \( 50 + 60 + 70 = 180 \). Correct. Order the angles. \( \angle A \lt \angle B \lt \angle C \). Match to opposite sides. \( \angle A \) faces \( \overline{BC} \), \( \angle B \) faces \( \overline{AC} \), \( \angle C \) faces \( \overline{AB} \). So the sides in increasing order are \( \overline{BC} \), \( \overline{AC} \), \( \overline{AB} \). Note the triangle is acute, since the largest angle is \( 70^\circ \), so no side is dramatically longer than the others. The ordering is still strict because all three angles differ. \( \overline{BC} \lt \overline{AC} \lt \overline{AB} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Can 3, 4, 5 form a triangle?
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   \( 3 + 4 = 7 \gt 5 \). Yes

2. Can 2, 3, 6 form a triangle?
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   \( 2 + 3 = 5 \lt 6 \). No

3. Can 5, 5, 10 form a triangle?
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   \( 5 + 5 = 10 \), not greater. Degenerate. No

4. Given sides 6 and 10, what is the upper bound on the third?
   Show the full solution

   Less than 16

5. Given sides 6 and 10, what is the lower bound?
   Show the full solution

   \( |10 - 6| \). Greater than 4

6. Two triangles have sides 8 and 12 with included angles \( 40^\circ \) and \( 95^\circ \). Which has the longer third side?
   Show the full solution

   By the hinge theorem, with two pairs of congruent sides, the larger included angle gives the longer third side. Since \( 95^\circ \gt 40^\circ \), the \( 95^\circ \) triangle has the longer third side. Verifying with the law of cosines: \( c^2 = 64 + 144 - 192\cos\theta \). At \( 40^\circ \): \( c^2 \approx 208 - 147.1 = 60.9 \), so \( c \approx 7.80 \). At \( 95^\circ \): \( c^2 \approx 208 + 16.7 = 224.7 \), so \( c \approx 14.99 \). Confirmed, and the difference is large. The one with the \( 95^\circ \) angle

7. A triangle has sides 9 and \( x \) with the third side 14. Find the range of \( x \).
   Show the full solution

   All three inequalities must hold. \( 9 + x \gt 14 \) gives \( x \gt 5 \). \( 9 + 14 \gt x \) gives \( x \lt 23 \). \( x + 14 \gt 9 \) gives \( x \gt -5 \), automatic. So \( 5 \lt x \lt 23 \). Checking against the shortcut: the two known sides are 9 and 14, so the range is \( |14 - 9| \lt x \lt 14 + 9 \), that is \( 5 \lt x \lt 23 \). Agrees. \( 5 \lt x \lt 23 \)

8. Two triangles have two pairs of congruent sides, with third sides of 11 and 15. Compare the included angles.
   Show the full solution

   This is the converse of the hinge theorem: the third sides are known and the angles are wanted. The larger third side belongs to the triangle with the larger included angle, so the triangle with third side 15 has the larger included angle. The reasoning mirrors the hinge theorem's door picture: opening the hinge wider increases the distance across, so a greater distance across means the hinge was opened wider. The triangle with third side 15 has the larger included angle

9. Explain why the triangle inequality follows from the fact that a segment is the shortest path.
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   Take three points \( A \), \( B \), \( C \) forming a triangle. Consider two routes from \( A \) to \( C \): directly along \( \overline{AC} \), or via \( B \) along \( \overline{AB} \) and then \( \overline{BC} \). The direct route has length \( AC \). The detour has length \( AB + BC \). Since a segment is the shortest path between two points, the direct route cannot be longer than any other route, so \( AC \leq AB + BC \). Equality would require the detour to be no longer than the direct path, which happens only if \( B \) lies on \( \overline{AC} \). In that case the three points are collinear and there is no triangle. For a genuine triangle \( B \) is off the line, so the inequality is strict: \( AC \lt AB + BC \). The same argument applied to the other two vertices gives the other two inequalities, and together they are the triangle inequality. The direct path from \( A \) to \( C \) cannot exceed the detour through \( B \), and it is strictly shorter unless the points are collinear

10. A triangle has sides \( x \), \( x + 3 \) and \( 2x - 1 \). Find all values of \( x \) for which this is a valid triangle.
    Show the full solution

    First, all sides must be positive. \( x \gt 0 \); \( x + 3 \gt 0 \) is automatic; \( 2x - 1 \gt 0 \) gives \( x \gt 0.5 \). So far \( x \gt 0.5 \). Now the three triangle inequalities. \( x + (x + 3) \gt 2x - 1 \): this gives \( 2x + 3 \gt 2x - 1 \), so \( 3 \gt -1 \), which is always true and imposes no constraint. \( x + (2x - 1) \gt x + 3 \): this gives \( 3x - 1 \gt x + 3 \), so \( 2x \gt 4 \) and \( x \gt 2 \). \( (x + 3) + (2x - 1) \gt x \): this gives \( 3x + 2 \gt x \), so \( 2x \gt -2 \) and \( x \gt -1 \), automatic. Combine. The binding constraint is \( x \gt 2 \), which also implies \( x \gt 0.5 \). Check a value inside the range. At \( x = 3 \) the sides are 3, 6 and 5. Checking: \( 3 + 5 = 8 \gt 6 \). Valid. Check a value outside. At \( x = 2 \) the sides are 2, 5 and 3, and \( 2 + 3 = 5 \), exactly equal to the third side. Degenerate, which is why the inequality is strict and \( x = 2 \) is excluded. At \( x = 1.5 \) the sides are 1.5, 4.5 and 2, and \( 1.5 + 2 = 3.5 \lt 4.5 \). Invalid, as predicted. \( x \gt 2 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the interior angle sum of a pentagon.
   Show the full solution

   \( (5-2) \times 180 \). \( 540^\circ \)

2. Find the interior angle sum of a decagon.
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   \( (10-2) \times 180 \). \( 1440^\circ \)

3. What is the exterior angle sum of any convex polygon?
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   \( 360^\circ \)

4. Find each interior angle of a regular hexagon.
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   \( \dfrac{(6-2)180}{6} = \dfrac{720}{6} \). \( 120^\circ \)

5. Find each exterior angle of a regular hexagon.
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   \( 360 \div 6 \). Check: \( 120 + 60 = 180 \). \( 60^\circ \)

6. A regular polygon has exterior angles of \( 24^\circ \). Find \( n \) and each interior angle.
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   The exterior angles sum to \( 360^\circ \), so \( n = \dfrac{360}{24} = 15 \). Each interior angle is the supplement: \( 180 - 24 = 156^\circ \). Check with the interior formula: \( \dfrac{(15-2)180}{15} = \dfrac{2340}{15} = 156 \). Agrees. \( n = 15 \), interior angles \( 156^\circ \)

7. A polygon's interior angles sum to \( 2340^\circ \). How many sides?
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   \( (n - 2) \cdot 180 = 2340 \), so \( n - 2 = \dfrac{2340}{180} = 13 \) and \( n = 15 \). Check: \( (15 - 2) \times 180 = 2340 \). Correct. Note this is the same polygon as the previous problem, reached from a different given. 15 sides

8. Four angles of a pentagon measure \( 100^\circ \), \( 110^\circ \), \( 120^\circ \) and \( 95^\circ \). Find the fifth.
   Show the full solution

   The interior angles of a pentagon sum to \( (5 - 2) \times 180 = 540^\circ \). The four given sum to \( 100 + 110 + 120 + 95 = 425 \). The fifth measures \( 540 - 425 = 115^\circ \). Check: all five are less than \( 180^\circ \), so the pentagon is convex and the formula applies as used. \( 115^\circ \)

9. Explain why the exterior angle sum is \( 360^\circ \) regardless of the number of sides.
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   Imagine walking once around the outside of the polygon, following each side and turning at each vertex before continuing along the next side. The amount you turn at a vertex is exactly the exterior angle there, since the exterior angle measures how far the direction of travel changes. By the time you return to the start facing the original direction, you have turned through one complete revolution, which is \( 360^\circ \). So the exterior angles sum to \( 360^\circ \). Nothing in that argument mentioned the number of sides. A triangle turns through three large angles; a hundred-gon turns through a hundred tiny ones; either way the total is one full turn. Algebraic confirmation: each interior and exterior pair sums to \( 180^\circ \), so all \( n \) pairs sum to \( 180n \). Subtracting the interior total: \( 180n - (n-2)180 = 180n - 180n + 360 = 360 \). The \( n \) terms cancel, which is precisely why the answer does not depend on \( n \). Walking around the polygon turns through exactly one full revolution, and the algebra shows the \( n \) terms cancel

10. The interior angles of a hexagon are in the ratio 3 : 4 : 5 : 5 : 6 : 7. Find each, and check whether the hexagon is convex.
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    Set up. Let the angles be \( 3x \), \( 4x \), \( 5x \), \( 5x \), \( 6x \) and \( 7x \). A hexagon's interior angles sum to \( (6 - 2) \times 180 = 720^\circ \). Solve. \( 3x + 4x + 5x + 5x + 6x + 7x = 720 \), so \( 30x = 720 \) and \( x = 24 \). The angles. \( 3(24) = 72^\circ \); \( 4(24) = 96^\circ \); \( 5(24) = 120^\circ \); \( 5(24) = 120^\circ \); \( 6(24) = 144^\circ \); \( 7(24) = 168^\circ \). Check the sum. \( 72 + 96 + 120 + 120 + 144 + 168 = 720 \). Correct. Check convexity. A polygon is convex when every interior angle is less than \( 180^\circ \). The largest here is \( 168^\circ \), so all six qualify and the hexagon is convex. That matters because the formula \( (n-2) \cdot 180 \) was stated for convex polygons. Had an angle come out above \( 180^\circ \), the answer would still be arithmetically consistent but the figure would be concave, and the exterior angle sum of \( 360^\circ \) would no longer apply in the same way. \( 72^\circ, 96^\circ, 120^\circ, 120^\circ, 144^\circ, 168^\circ \); convex, since all are under \( 180^\circ \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is the definition of a parallelogram?
   Show the full solution

   A quadrilateral with both pairs of opposite sides parallel

2. In parallelogram \( ABCD \), \( m\angle A = 72^\circ \). Find \( m\angle C \).
   Show the full solution

   Opposite angles are congruent. \( 72^\circ \)

3. Find \( m\angle B \) in that parallelogram.
   Show the full solution

   Consecutive angles are supplementary. \( 108^\circ \)

4. In parallelogram \( ABCD \), \( AB = 9 \). Find \( CD \).
   Show the full solution

   9

5. Are the diagonals of a parallelogram congruent?
   Show the full solution

   They bisect each other, which is different. Not in general

6. In parallelogram \( ABCD \), the diagonals meet at \( E \) with \( AE = 3x - 1 \) and \( EC = x + 7 \). Find \( AC \).
   Show the full solution

   The diagonals bisect each other, so \( AE = EC \): \( 3x - 1 = x + 7 \), giving \( 2x = 8 \) and \( x = 4 \). Then \( AE = 3(4) - 1 = 11 \) and \( EC = 4 + 7 = 11 \). Equal, as required. \( AC = AE + EC = 11 + 11 = 22 \). The question asked for the whole diagonal, not the half, which is the step most often skipped. \( AC = 22 \)

7. In parallelogram \( PQRS \), \( m\angle P = (2x + 10)^\circ \) and \( m\angle Q = (3x - 5)^\circ \). Find all four angles.
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   \( \angle P \) and \( \angle Q \) are consecutive, so they are supplementary: \( (2x + 10) + (3x - 5) = 180 \), giving \( 5x + 5 = 180 \), so \( 5x = 175 \) and \( x = 35 \). \( m\angle P = 2(35) + 10 = 80^\circ \) and \( m\angle Q = 3(35) - 5 = 100^\circ \). Check: \( 80 + 100 = 180 \). Correct. Opposite angles: \( m\angle R = m\angle P = 80^\circ \) and \( m\angle S = m\angle Q = 100^\circ \). Total: \( 80 + 100 + 80 + 100 = 360 \). Correct. \( 80^\circ \), \( 100^\circ \), \( 80^\circ \), \( 100^\circ \)

8. Prove that opposite angles of a parallelogram are congruent.
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   Given: parallelogram \( ABCD \). Prove: \( \angle A \cong \angle C \). Route one, by triangles. 1. \( ABCD \) is a parallelogram. Reason: given. 2. Draw \( \overline{BD} \). Reason: two points determine a segment. 3. \( \angle ABD \cong \angle CDB \) and \( \angle ADB \cong \angle CBD \). Reason: alternate interior angles theorem, applied to each pair of parallel sides. 4. \( \overline{BD} \cong \overline{BD} \). Reason: reflexive property of congruence. 5. \( \triangle ABD \cong \triangle CDB \). Reason: ASA. 6. \( \angle A \cong \angle C \). Reason: CPCTC. Route two, by supplements. \( \angle A \) and \( \angle B \) are supplementary, and \( \angle B \) and \( \angle C \) are supplementary, both as consecutive angles. Two angles supplementary to the same angle are congruent by the congruent supplements theorem of lesson 2.6, so \( \angle A \cong \angle C \). The second route is shorter and uses the consecutive angle property already proved. Proved, by ASA and CPCTC or by congruent supplements

9. Explain why every parallelogram property is a theorem rather than part of the definition.
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   A definition should say the least that identifies the object. Once "both pairs of opposite sides parallel" is stated, the figure is completely determined as a type, and everything else about it follows. Including the other properties in the definition would be redundant and, worse, would hide the logical structure. A reader would not know which facts are independent and which follow, and a proof could not be checked for circularity. There is also a practical benefit. Because the properties are theorems, each can be cited by name in later proofs without rebuilding the triangle argument. Proving "opposite sides are congruent" once means every subsequent proof gets it in one line. The same principle governs the rest of the unit. A rectangle is defined as a parallelogram with one additional condition, and its diagonal property is then a theorem proved from that. A definition names; theorems describe. A definition should state the minimum that identifies the figure; everything derivable becomes a theorem, which keeps the logical structure visible and lets each result be cited by name

10. In parallelogram \( ABCD \), \( AB = 2x + 5 \), \( BC = 3y - 4 \), \( CD = 17 \), \( AD = 11 \). Find \( x \), \( y \) and the perimeter.
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    Opposite sides of a parallelogram are congruent, so pair them correctly: \( \overline{AB} \) is opposite \( \overline{CD} \), and \( \overline{BC} \) is opposite \( \overline{AD} \). Find \( x \). \( AB = CD \) gives \( 2x + 5 = 17 \), so \( 2x = 12 \) and \( x = 6 \). Then \( AB = 17 \). Find \( y \). \( BC = AD \) gives \( 3y - 4 = 11 \), so \( 3y = 15 \) and \( y = 5 \). Then \( BC = 11 \). Perimeter. The four sides are 17, 11, 17 and 11, so the perimeter is \( 2(17) + 2(11) = 34 + 22 = 56 \). Checks. Both computed values are positive, as lengths must be. The perimeter formula \( 2(a + b) \) for a parallelogram gives \( 2(17 + 11) = 56 \), agreeing. The trap here is pairing \( \overline{AB} \) with \( \overline{AD} \), which are adjacent rather than opposite and need not be congruent. Reading the vertex letters rather than the order of presentation is what prevents it. \( x = 6 \), \( y = 5 \), perimeter 56

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Name the five tests for a parallelogram.
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   Both pairs of sides parallel; both pairs congruent; both pairs of angles congruent; diagonals bisect each other; one pair both parallel and congruent

2. A quadrilateral has diagonals that bisect each other. Is it a parallelogram?
   Show the full solution

   Yes

3. A quadrilateral has one pair of parallel sides. Is it a parallelogram?
   Show the full solution

   That is a trapezoid. Not necessarily

4. A quadrilateral has both pairs of opposite sides congruent. Is it a parallelogram?
   Show the full solution

   Yes

5. What is the difference between a property and a test?
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   A property follows from being a parallelogram; a test proves something is one

6. A quadrilateral has \( \overline{AB} \parallel \overline{DC} \) and \( \overline{AB} \cong \overline{DC} \). Which test applies?
   Show the full solution

   Both conditions are on the same pair of opposite sides. Test five

7. In quadrilateral \( ABCD \), \( m\angle A = 110^\circ \), \( m\angle B = 70^\circ \), \( m\angle C = 110^\circ \). Find \( m\angle D \) and decide whether it is a parallelogram.
   Show the full solution

   The angles of a quadrilateral sum to \( 360^\circ \), so \( m\angle D = 360 - 110 - 70 - 110 = 70^\circ \). Now check the opposite pairs: \( \angle A \) and \( \angle C \) are both \( 110^\circ \), congruent; \( \angle B \) and \( \angle D \) are both \( 70^\circ \), congruent. Both pairs of opposite angles are congruent, which is test three. \( m\angle D = 70^\circ \), and yes, it is a parallelogram

8. Prove test four: if the diagonals of a quadrilateral bisect each other, it is a parallelogram.
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   Given: quadrilateral \( ABCD \) with diagonals meeting at \( E \), and \( \overline{AE} \cong \overline{EC} \), \( \overline{BE} \cong \overline{ED} \). Prove: \( ABCD \) is a parallelogram. 1. \( \overline{AE} \cong \overline{EC} \); \( \overline{BE} \cong \overline{ED} \). Reason: given. 2. \( \angle AEB \cong \angle CED \). Reason: vertical angles theorem. 3. \( \triangle AEB \cong \triangle CED \). Reason: SAS. 4. \( \overline{AB} \cong \overline{CD} \). Reason: CPCTC. 5. \( \angle BAE \cong \angle DCE \). Reason: CPCTC. 6. \( \overline{AB} \parallel \overline{DC} \). Reason: converse of the alternate interior angles theorem, using \( \overline{AC} \) as transversal. 7. \( ABCD \) is a parallelogram. Reason: test five, since \( \overline{AB} \) and \( \overline{DC} \) are both parallel and congruent. An alternative ending repeats the argument on \( \triangle AED \) and \( \triangle CEB \) to get the second pair of sides congruent and then cites test two. Either is valid. Proved by SAS, CPCTC and test five

9. Give a counterexample showing that one pair of congruent sides alone does not make a parallelogram.
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   Take the quadrilateral with vertices \( A(0,0) \), \( B(6,0) \), \( C(7,3) \), \( D(1,4) \). Check the side lengths: \( AB = 6 \); \( BC = \sqrt{1 + 9} = \sqrt{10} \); \( CD = \sqrt{36 + 1} = \sqrt{37} \); \( DA = \sqrt{1 + 16} = \sqrt{17} \). None are congruent, so amend the example: take \( A(0,0) \), \( B(6,0) \), \( C(8,3) \), \( D(2,5) \). \( AB = 6 \); \( CD = \sqrt{36 + 4} = \sqrt{40} \); \( BC = \sqrt{4 + 9} = \sqrt{13} \); \( DA = \sqrt{4 + 25} = \sqrt{29} \). Still none congruent. A cleaner construction: a kite. Take \( A(0,0) \), \( B(3,4) \), \( C(6,0) \), \( D(3,-2) \). \( AB = \sqrt{9 + 16} = 5 \) and \( BC = \sqrt{9 + 16} = 5 \), so those two adjacent sides are congruent. \( AD = \sqrt{9 + 4} = \sqrt{13} \) and \( CD = \sqrt{9 + 4} = \sqrt{13} \), also congruent. This figure has two pairs of congruent sides, but they are adjacent pairs rather than opposite pairs. Its opposite sides are \( AB = 5 \) against \( CD = \sqrt{13} \), not congruent, so it is not a parallelogram. It is a kite. The lesson: congruent sides must be opposite pairs for test two to apply, and a figure can have plenty of congruent sides arranged the wrong way. A kite, such as \( (0,0) \), \( (3,4) \), \( (6,0) \), \( (3,-2) \), has two pairs of congruent adjacent sides and is not a parallelogram

10. In quadrilateral \( ABCD \), \( \overline{AB} \cong \overline{CD} \) and \( \overline{AD} \parallel \overline{BC} \). Determine whether it must be a parallelogram, and give a complete argument.
    Show the full solution

    The given facts are a congruent pair and a parallel pair, but on different pairs of opposite sides. This is exactly the configuration the worked example showed to be insufficient. The counterexample. Take the isosceles trapezoid with \( A(0,0) \), \( B(6,0) \), \( C(5,3) \), \( D(1,3) \), relabeled so the conditions match. Here \( \overline{AD} \) from \( (0,0) \) to \( (1,3) \) has length \( \sqrt{10} \), and \( \overline{BC} \) from \( (6,0) \) to \( (5,3) \) has length \( \sqrt{10} \), so those are congruent, while \( \overline{AB} \) and \( \overline{DC} \) are parallel, both horizontal. Renaming to match the problem's labels: the figure has one pair of opposite sides congruent and the other pair parallel, and it is not a parallelogram, since \( AB = 6 \) and \( DC = 4 \) differ. So the answer is no, it need not be a parallelogram. What it could be. Either a parallelogram or an isosceles trapezoid. Those are the only two possibilities: the parallel pair makes it at least a trapezoid, and the congruent legs make it isosceles unless the legs are also parallel, in which case it is a parallelogram. What would settle it. Knowing that \( \overline{AD} \) and \( \overline{BC} \) are also parallel, or that \( \overline{AB} \cong \overline{DC} \), or that the diagonals bisect each other. Any one of those converts the figure into a parallelogram by one of the five tests. This is the same trap as the "where students lose marks" note, and it is worth recognizing by shape: conditions spread across two different pairs of sides are almost never enough. No; an isosceles trapezoid satisfies both conditions without being a parallelogram

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What extra condition makes a parallelogram a rectangle?
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   A right angle

2. What extra condition makes a parallelogram a rhombus?
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   Two adjacent congruent sides

3. Which figure has congruent diagonals?
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   The rectangle, and therefore the square

4. Which figure has perpendicular diagonals?
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   The rhombus, and therefore the square

5. A rhombus has diagonals of 6 and 8. Find its side.
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   Halves are 3 and 4, so \( s = \sqrt{9 + 16} = 5 \). 5

6. A rectangle has sides 9 and 12. Find the length of each diagonal.
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   A diagonal is the hypotenuse of a right triangle with legs equal to the sides, since a rectangle has right angles. \( d = \sqrt{81 + 144} = \sqrt{225} = 15 \). Both diagonals measure 15, since the diagonals of a rectangle are congruent. 15 each

7. A square has a diagonal of \( 10\sqrt{2} \). Find its side and area.
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   In a square the diagonal is the hypotenuse of a right triangle whose legs are two sides, both of length \( s \): \( d = \sqrt{s^2 + s^2} = s\sqrt{2} \). So \( s\sqrt{2} = 10\sqrt{2} \), giving \( s = 10 \). Area: \( s^2 = 100 \). Check with the rhombus area formula, since a square is a rhombus: half the product of the diagonals is \( \dfrac{1}{2}(10\sqrt{2})(10\sqrt{2}) = \dfrac{1}{2}(200) = 100 \). Agrees. Side 10, area 100

8. A parallelogram has one right angle. Prove it is a rectangle.
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   Given: parallelogram \( ABCD \) with \( \angle A \) a right angle. Prove: \( ABCD \) is a rectangle. 1. \( ABCD \) is a parallelogram with \( m\angle A = 90 \). Reason: given. 2. \( m\angle C = m\angle A = 90 \). Reason: opposite angles of a parallelogram are congruent. 3. \( \angle A \) and \( \angle B \) are supplementary. Reason: consecutive angles of a parallelogram are supplementary. 4. \( m\angle B = 180 - 90 = 90 \). Reason: definition of supplementary angles and substitution. 5. \( m\angle D = m\angle B = 90 \). Reason: opposite angles of a parallelogram are congruent. 6. All four angles are right angles. Reason: lines 1, 2, 4 and 5 with the definition of a right angle. 7. \( ABCD \) is a rectangle. Reason: definition of a rectangle. This is why the definition needs only one right angle: the parallelogram properties propagate it to all four. Proved in seven lines

9. Explain why every square is a rectangle and a rhombus, but not every rectangle is a square.
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   A square is defined as having four right angles and four congruent sides. A rectangle is a parallelogram with four right angles. A square has four right angles, and it is a parallelogram since its opposite sides are parallel, so it satisfies the rectangle definition. Every square is a rectangle. A rhombus is a parallelogram with four congruent sides. A square has four congruent sides, so it satisfies that definition too. Every square is a rhombus. The converse fails in both directions because each of those definitions demands less. A 3 by 5 rectangle has four right angles and is not a square, because its sides are not all congruent. A rhombus with \( 60^\circ \) and \( 120^\circ \) angles has four congruent sides and is not a square, because its angles are not right. The general pattern of this unit: adding a condition narrows the class, so the more specific figure is always a member of the broader one and not conversely. Lesson 7.6 organizes all of this into a hierarchy. A square satisfies both definitions by having both properties; a rectangle lacks the congruent sides and a rhombus lacks the right angles

10. A rhombus has a side of 13 and one diagonal of 24. Find the other diagonal and the area.
    Show the full solution

    Set up. The diagonals of a rhombus are perpendicular and bisect each other, so they cut it into four congruent right triangles whose legs are the half diagonals and whose hypotenuse is a side. Half of the known diagonal is \( \dfrac{24}{2} = 12 \). Find the other half diagonal. By the Pythagorean theorem, \( 12^2 + b^2 = 13^2 \), so \( b^2 = 169 - 144 = 25 \) and \( b = 5 \). The other diagonal is \( 2 \times 5 = 10 \). Find the area. For a rhombus the area is half the product of the diagonals: \( \dfrac{1}{2}(24)(10) = 120 \). Check by a second method. Four congruent right triangles with legs 12 and 5: \( 4 \times \dfrac{1}{2}(12)(5) = 4 \times 30 = 120 \). Agrees. Check by a third method. A rhombus is a parallelogram, so its area is base times height. The base is 13. The height can be found from the area: \( h = \dfrac{120}{13} \approx 9.23 \), which is less than the side of 13 as a height must be. Consistent. Sanity check on the numbers. The side 13 must exceed each half diagonal, and \( 13 \gt 12 \gt 5 \). Correct. The 5-12-13 right triangle is the standard one, which is why the numbers came out whole. The other diagonal is 10 and the area is 120

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is the midsegment of a trapezoid with bases 6 and 10?
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   Average them. 8

2. What makes a trapezoid isosceles?
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   Congruent legs

3. Are the diagonals of an isosceles trapezoid congruent?
   Show the full solution

   Yes

4. What are the sides of a kite like?
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   Two distinct pairs of congruent adjacent sides

5. Are the diagonals of a kite perpendicular?
   Show the full solution

   Yes

6. A trapezoid has bases \( 3x + 1 \) and \( x + 9 \) with midsegment 15. Find both bases.
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   The midsegment averages the bases: \( \dfrac{(3x + 1) + (x + 9)}{2} = 15 \), so \( 4x + 10 = 30 \), giving \( 4x = 20 \) and \( x = 5 \). Bases: \( 3(5) + 1 = 16 \) and \( 5 + 9 = 14 \). Check: \( \dfrac{16 + 14}{2} = 15 \). Correct, and 15 lies between 14 and 16. 16 and 14

7. An isosceles trapezoid has a base angle of \( 62^\circ \). Find all four angles.
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   Base angles come in congruent pairs, so two angles measure \( 62^\circ \) each. The legs are transversals of the two parallel bases, so each pair of angles on the same leg is a same-side interior pair and therefore supplementary: \( 180 - 62 = 118^\circ \). So the other two angles measure \( 118^\circ \) each. Check: \( 62 + 62 + 118 + 118 = 360 \), the quadrilateral angle sum. Correct. \( 62^\circ \), \( 62^\circ \), \( 118^\circ \), \( 118^\circ \)

8. A kite has sides of 7, 7, 13 and 13. Its shorter diagonal is 10. Find the longer diagonal.
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   The shorter diagonal joins the two vertices where the unequal sides meet, and the longer diagonal bisects it, so the halves are 5 each. The longer diagonal is made of two pieces, one in each of the two triangles the shorter diagonal creates. In the triangle with two sides of 7: the piece is \( \sqrt{49 - 25} = \sqrt{24} = 2\sqrt{6} \approx 4.90 \). In the triangle with two sides of 13: the piece is \( \sqrt{169 - 25} = \sqrt{144} = 12 \). Total: \( 12 + 2\sqrt{6} \approx 16.90 \). Check: the diagonals are perpendicular, so the area is \( \dfrac{1}{2}(10)(12 + 2\sqrt{6}) \approx \dfrac{1}{2}(10)(16.90) \approx 84.5 \), a plausible area for a figure with these side lengths. \( 12 + 2\sqrt{6} \approx 16.90 \)

9. Prove the trapezoid midsegment theorem using the triangle midsegment theorem.
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   Given: trapezoid \( ABCD \) with \( \overline{AB} \parallel \overline{DC} \), \( M \) the midpoint of leg \( \overline{AD} \), \( N \) the midpoint of leg \( \overline{BC} \). Prove: \( MN = \dfrac{AB + DC}{2} \). The construction. Draw the diagonal \( \overline{AC} \), meeting \( \overline{MN} \) at a point \( P \). In \( \triangle ADC \). \( M \) is the midpoint of \( \overline{AD} \), and \( \overline{MP} \) is parallel to \( \overline{DC} \), since \( \overline{MN} \) is parallel to both bases. By the converse of the triangle midsegment theorem, \( P \) is the midpoint of \( \overline{AC} \), and \( \overline{MP} \) is a midsegment, so \( MP = \dfrac{DC}{2} \). In \( \triangle ABC \). \( P \) is the midpoint of \( \overline{AC} \) and \( N \) is the midpoint of \( \overline{BC} \), so \( \overline{PN} \) is a midsegment and \( PN = \dfrac{AB}{2} \). Combine. \( MN = MP + PN = \dfrac{DC}{2} + \dfrac{AB}{2} = \dfrac{AB + DC}{2} \). So the trapezoid midsegment is the average of the bases, proved by splitting it into two triangle midsegments. Proved by drawing a diagonal and applying the triangle midsegment theorem twice

10. An isosceles trapezoid has bases 10 and 22 and legs of 10. Find its height and area.
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    Set up. Drop altitudes from the two endpoints of the shorter base to the longer base. They cut the longer base into three pieces: two equal end pieces and a middle piece equal to the shorter base. The middle piece is 10, so the two end pieces together measure \( 22 - 10 = 12 \), and each is \( \dfrac{12}{2} = 6 \). The end pieces are equal because the trapezoid is isosceles, which is where that condition is used. Find the height. Each leg is the hypotenuse of a right triangle with horizontal leg 6 and vertical leg \( h \): \( 6^2 + h^2 = 10^2 \), so \( h^2 = 100 - 36 = 64 \) and \( h = 8 \). Find the area. A trapezoid's area is the midsegment times the height, or equivalently \( \dfrac{b_1 + b_2}{2} \cdot h \): \( \dfrac{10 + 22}{2} \times 8 = 16 \times 8 = 128 \). Check by decomposition. The figure is a 10 by 8 rectangle in the middle plus two right triangles with legs 6 and 8 at the ends: \( 80 + 2 \times \dfrac{1}{2}(6)(8) = 80 + 48 = 128 \). Agrees. Sanity check. The leg of 10 is the hypotenuse of a 6-8-10 triangle, so the height 8 is less than the leg 10 as it must be. And the midsegment is 16, between the bases of 10 and 22. Both consistent. Height 8, area 128

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Is every square a rectangle?
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   Yes

2. Is every rectangle a square?
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   No

3. Is every rhombus a parallelogram?
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   Yes

4. Is a trapezoid a parallelogram?
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   Under the exclusive definition used here. No

5. Name the most specific figure with four congruent sides and four right angles.
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   Square

6. A parallelogram has two congruent adjacent sides. Classify it as specifically as possible.
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   In a parallelogram, opposite sides are already congruent. Adding that two adjacent sides are congruent makes all four congruent, by the transitive property. A parallelogram with four congruent sides is a rhombus. It is not necessarily a square, since nothing about the angles is given. A rhombus with \( 70^\circ \) and \( 110^\circ \) angles satisfies the conditions and is not a square. A rhombus

7. A quadrilateral has congruent diagonals. Must it be a rectangle?
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   No. The theorem is that a parallelogram with congruent diagonals is a rectangle, and the parallelogram condition is doing real work. A counterexample: an isosceles trapezoid has congruent diagonals, as lesson 7.5 established, and it is not a rectangle. It is not even a parallelogram. What would be needed: the diagonals must also bisect each other, which establishes the parallelogram, and then congruence gives the rectangle. This is a good example of the general error of dropping a hypothesis. The conclusion of a theorem depends on all of its conditions, and a question that supplies only some of them supports only a weaker conclusion. No; an isosceles trapezoid is a counterexample

8. List every name that applies to a square, from most to least specific.
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   Square, rectangle, rhombus, parallelogram, quadrilateral, polygon. Each name up the list is broader and each is genuinely correct. A square really is a rectangle, because it satisfies the rectangle definition of a parallelogram with four right angles, and it really is a rhombus, because it has four congruent sides. What a classification question wants is the first name on this list, the most specific justified one. Answering "parallelogram" is not false but is marked down, because a more specific name was available and supported. Note that rectangle and rhombus are at the same level rather than one above the other. Neither implies the other, and the square is exactly their intersection. Square, rectangle, rhombus, parallelogram, quadrilateral, polygon

9. Explain why the exclusive definition of a trapezoid puts it outside the parallelogram branch, and what changes under the inclusive definition.
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   Exclusive definition: a trapezoid has exactly one pair of parallel sides. A parallelogram has two pairs, so it fails the "exactly one" condition and is not a trapezoid. The two branches are disjoint, and the hierarchy has trapezoids and kites as separate families from parallelograms. Inclusive definition: a trapezoid has at least one pair of parallel sides. Then every parallelogram qualifies as a trapezoid, and the parallelogram branch sits underneath the trapezoid branch. The hierarchy becomes a single tree with trapezoid above parallelogram. What changes in practice. Under the inclusive definition, "this is a trapezoid" is true of a square, which sounds wrong to most students and is why many courses prefer the exclusive version. But the inclusive version makes the trapezoid midsegment formula apply to parallelograms as a special case: with \( b_1 = b_2 \), the average is that common value, which is correct for a parallelogram's midsegment. What to do on a test. Use the definition the course states, and if a question turns on the distinction, say which definition is being used. The mathematics is identical; only the naming convention differs. Exclusive keeps the branches separate; inclusive makes every parallelogram a trapezoid and unifies the midsegment formula

10. A quadrilateral has diagonals that are perpendicular. List everything it could be, and what extra information would narrow it each way.
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    Perpendicular diagonals alone are a weak condition. Three named families have them. A rhombus. Its diagonals are perpendicular and also bisect each other. To narrow to this, add that the diagonals bisect each other, which gives a parallelogram, and combined with perpendicularity gives a rhombus. A square. A special rhombus. To narrow to this, add that the diagonals are also congruent and bisect each other. A kite. Its diagonals are perpendicular, and one bisects the other but not conversely. To narrow to this, add that exactly one diagonal bisects the other, or that there are two pairs of congruent adjacent sides. And an unnamed quadrilateral. Two segments crossing at right angles with neither bisecting the other still determine a quadrilateral, and it has no special name. Take diagonals crossing so that one is split 2 and 5 and the other 1 and 7: the four sides are all different and no family applies. The general lesson. A single diagonal condition rarely classifies a figure. The useful classifier is the combination: bisecting gives parallelogram, adding congruent gives rectangle, adding perpendicular gives rhombus, adding both gives square. Perpendicular without bisecting leaves the kite and the general case open. A rhombus, a square, a kite, or an unnamed quadrilateral; adding that the diagonals bisect each other narrows it to rhombus, and adding congruence as well gives a square

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Which formula tests for parallel sides?
   Show the full solution

   Slope

2. Which formula tests for congruent sides?
   Show the full solution

   Distance

3. Which formula tests whether diagonals bisect each other?
   Show the full solution

   Midpoint; both diagonals must share one

4. Two adjacent sides have slopes 2 and \( -\frac{1}{2} \). What follows?
   Show the full solution

   The product is \( -1 \). They are perpendicular, so that angle is right

5. What is the most specific figure with four congruent sides and a right angle?
   Show the full solution

   A square

6. Classify the quadrilateral \( (0,0) \), \( (5,0) \), \( (7,4) \), \( (2,4) \).
   Show the full solution

   Slopes. Bottom: \( \dfrac{0-0}{5-0} = 0 \). Right: \( \dfrac{4-0}{7-5} = 2 \). Top: \( \dfrac{4-4}{2-7} = 0 \). Left: \( \dfrac{0-4}{0-2} = 2 \). Both pairs of opposite sides are parallel, so it is a parallelogram. Lengths. Bottom: 5. Right: \( \sqrt{4 + 16} = \sqrt{20} \). Top: 5. Left: \( \sqrt{4 + 16} = \sqrt{20} \). Opposite sides congruent, as a parallelogram requires, but adjacent sides are not congruent since \( 5 \neq \sqrt{20} \approx 4.47 \). So it is not a rhombus. Right angles. \( 0 \times 2 = 0 \), not \( -1 \), so adjacent sides are not perpendicular. Not a rectangle. A parallelogram, and nothing more specific

7. Classify the quadrilateral \( (0,0) \), \( (4,0) \), \( (4,4) \), \( (0,4) \).
   Show the full solution

   Slopes. Bottom 0, right undefined (vertical), top 0, left undefined. Both pairs of opposite sides parallel, so a parallelogram. Right angles. A horizontal side meets a vertical side, which are perpendicular. So all four angles are right and it is a rectangle. Lengths. All four sides measure 4, so all are congruent and it is also a rhombus. A figure that is both a rectangle and a rhombus is a square. Diagonals as a check. From \( (0,0) \) to \( (4,4) \): \( \sqrt{32} = 4\sqrt{2} \). From \( (4,0) \) to \( (0,4) \): \( \sqrt{32} = 4\sqrt{2} \). Congruent, as a rectangle requires. Slopes 1 and \( -1 \), product \( -1 \), perpendicular as a rhombus requires. Both confirm. A square

8. Classify the quadrilateral \( (0,0) \), \( (6,0) \), \( (5,3) \), \( (1,3) \).
   Show the full solution

   Slopes. Bottom: 0. Right, from \( (6,0) \) to \( (5,3) \): \( \dfrac{3}{-1} = -3 \). Top, from \( (5,3) \) to \( (1,3) \): 0. Left, from \( (1,3) \) to \( (0,0) \): \( \dfrac{-3}{-1} = 3 \). The bottom and top are parallel, both slope 0. The left and right are not, since \( 3 \neq -3 \). Exactly one pair of parallel sides, so it is a trapezoid. Legs. Right: \( \sqrt{1 + 9} = \sqrt{10} \). Left: \( \sqrt{1 + 9} = \sqrt{10} \). Congruent, so it is an isosceles trapezoid. Diagonals as a check. From \( (0,0) \) to \( (5,3) \): \( \sqrt{34} \). From \( (6,0) \) to \( (1,3) \): \( \sqrt{25 + 9} = \sqrt{34} \). Congruent, as an isosceles trapezoid requires. An isosceles trapezoid

9. Explain why a coordinate classification counts as a proof.
   Show the full solution

   A proof establishes a conclusion from stated facts by named justifications, and a coordinate classification does exactly that. The given facts are the coordinates of the vertices. The justifications are the slope, distance and midpoint formulas, each of which is a theorem proved earlier in the course, together with the tests and definitions of this unit. A complete classification says: these two slopes are equal, so by the slope criterion these sides are parallel; both pairs are parallel, so by the definition the figure is a parallelogram; these four distances are equal, so by the definition it is a rhombus. Each line has a statement and a reason, exactly as a two-column proof does. What makes it fail as a proof is presenting the numbers without the conclusions. Computing four slopes and then writing "so it is a rhombus" skips every justification and is not an argument. There is also a generality question worth noting. A coordinate proof about a specific figure proves something about that figure only. To prove a general theorem by coordinates, as in the midsegment proof of lesson 6.5, the vertices must be given variable coordinates so the argument covers every case. Each computation cites a formula and supports a conclusion by a named test, so the chain from coordinates to classification is a genuine argument

10. Classify the quadrilateral \( (0,0) \), \( (3,4) \), \( (8,4) \), \( (5,0) \), then determine its area two ways.
    Show the full solution

    Slopes. \( (0,0) \) to \( (3,4) \): \( \dfrac{4}{3} \). \( (3,4) \) to \( (8,4) \): 0. \( (8,4) \) to \( (5,0) \): \( \dfrac{-4}{-3} = \dfrac{4}{3} \). \( (5,0) \) to \( (0,0) \): 0. Both pairs of opposite sides parallel, so it is a parallelogram. Lengths. \( \sqrt{9 + 16} = 5 \); 5; \( \sqrt{9 + 16} = 5 \); 5. All four sides are 5, so it is a rhombus. Right angles. \( \dfrac{4}{3} \times 0 = 0 \), not \( -1 \). Not a rectangle, so not a square. Classification: a rhombus. Area, method one: base times height. Take the bottom side from \( (0,0) \) to \( (5,0) \) as the base, length 5. The opposite side lies along \( y = 4 \), so the height is 4. Area: \( 5 \times 4 = 20 \). Area, method two: half the product of the diagonals. Diagonal from \( (0,0) \) to \( (8,4) \): \( \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \). Diagonal from \( (3,4) \) to \( (5,0) \): \( \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \). Area: \( \dfrac{1}{2}(4\sqrt{5})(2\sqrt{5}) = \dfrac{1}{2}(8 \times 5) = 20 \). The two methods agree at 20. The second method is available only because the figure is a rhombus with perpendicular diagonals, which the classification established. Checking: the diagonal slopes are \( \dfrac{4}{8} = \dfrac{1}{2} \) and \( \dfrac{-4}{2} = -2 \), whose product is \( -1 \). Perpendicular, as required. A rhombus, with area 20 by both methods

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( \dfrac{x}{6} = \dfrac{10}{15} \).
   Show the full solution

   \( 15x = 60 \). \( x = 4 \)

2. Solve \( \dfrac{7}{x} = \dfrac{21}{9} \).
   Show the full solution

   \( 21x = 63 \). \( x = 3 \)

3. What does cross multiplying \( \dfrac{a}{b} = \dfrac{c}{d} \) give?
   Show the full solution

   \( ad = bc \)

4. Two numbers are in the ratio \( 2 : 5 \) and sum to 42. Find them.
   Show the full solution

   \( 2x + 5x = 42 \), so \( x = 6 \). 12 and 30

5. Simplify the ratio \( 18 : 24 \).
   Show the full solution

   Divide both by 6. \( 3 : 4 \)

6. Solve \( \dfrac{2x - 1}{3} = \dfrac{x + 4}{2} \).
   Show the full solution

   Cross multiply: \( 2(2x - 1) = 3(x + 4) \), so \( 4x - 2 = 3x + 12 \), giving \( x = 14 \). Check: \( \dfrac{2(14) - 1}{3} = \dfrac{27}{3} = 9 \) and \( \dfrac{14 + 4}{2} = \dfrac{18}{2} = 9 \). Equal. \( x = 14 \)

7. The angles of a triangle are in the ratio \( 2 : 3 : 7 \). Find them and classify the triangle.
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   Let the angles be \( 2x \), \( 3x \), \( 7x \). They sum to \( 180^\circ \): \( 12x = 180 \), so \( x = 15 \). The angles are \( 30^\circ \), \( 45^\circ \) and \( 105^\circ \). Check: \( 30 + 45 + 105 = 180 \). Correct. Since one angle exceeds \( 90^\circ \), the triangle is obtuse. It is also scalene, since all three angles differ. \( 30^\circ \), \( 45^\circ \), \( 105^\circ \); obtuse scalene

8. The sides of a triangle are in the ratio \( 5 : 12 : 13 \) with perimeter 90. Find the sides and classify it.
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   \( 5x + 12x + 13x = 90 \), so \( 30x = 90 \) and \( x = 3 \). Sides: 15, 36, 39. Check the perimeter: \( 15 + 36 + 39 = 90 \). Correct. Classify with the converse of the Pythagorean theorem: \( 15^2 + 36^2 = 225 + 1296 = 1521 \) and \( 39^2 = 1521 \). Equal, so the triangle is right. Any triangle with sides in the ratio \( 5 : 12 : 13 \) is right, for the same reason as \( 3 : 4 : 5 \). 15, 36, 39; a right triangle

9. Explain why \( \dfrac{a}{b} = \dfrac{c}{d} \) gives \( \dfrac{a}{c} = \dfrac{b}{d} \) but not \( \dfrac{a}{d} = \dfrac{c}{b} \).
   Show the full solution

   Why the first works. Cross multiplying the original gives \( ad = bc \). Cross multiplying the proposed rearrangement \( \dfrac{a}{c} = \dfrac{b}{d} \) gives \( ad = cb \). Those are the same equation, since multiplication is commutative. So the two proportions are equivalent and each implies the other. Why the second fails. Cross multiplying \( \dfrac{a}{d} = \dfrac{c}{b} \) gives \( ab = cd \), which is a completely different equation from \( ad = bc \). Nothing connects them. The test. Cross multiply both the original and the proposed rearrangement. If they produce the same equation, the rearrangement is valid; if not, it is not. A counterexample for the invalid one. With \( a = 1 \), \( b = 2 \), \( c = 3 \), \( d = 6 \): the original holds since \( \dfrac{1}{2} = \dfrac{3}{6} \), but \( \dfrac{1}{6} \neq \dfrac{3}{2} \). Cross multiplying both gives the same equation for the valid rearrangement and a different one for the invalid

10. A rectangle has sides in the ratio \( 3 : 8 \) and area 384. Find its dimensions and perimeter.
    Show the full solution

    Set up. Let the sides be \( 3x \) and \( 8x \). Area: \( (3x)(8x) = 24x^2 \). Solve. \( 24x^2 = 384 \), so \( x^2 = 16 \) and \( x = \pm 4 \). A length cannot be negative, so \( x = 4 \). Dimensions. \( 3(4) = 12 \) and \( 8(4) = 32 \). Check the area. \( 12 \times 32 = 384 \). Correct. Check the ratio. \( 12 : 32 \) simplifies by dividing both by 4 to \( 3 : 8 \). Correct. Perimeter. \( 2(12 + 32) = 2(44) = 88 \). A point worth noting. The ratio of the sides is \( 3 : 8 \), but the area involved \( x^2 \) rather than \( x \), which is why the scale factor had to be found by taking a square root. That is the same phenomenon as the area scaling rule of lesson 11.6: lengths scale by \( k \) and areas by \( k^2 \), so recovering a length from an area requires a square root. 12 by 32, perimeter 88

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Dilate \( (3, 5) \) about the origin with \( k = 4 \).
   Show the full solution

   \( (12, 20) \)

2. Dilate \( (8, 12) \) about the origin with \( k = \frac{1}{4} \).
   Show the full solution

   \( (2, 3) \)

3. Is a dilation a rigid motion?
   Show the full solution

   No, unless \( k = 1 \)

4. A dilation has \( k = 0.5 \). Enlargement or reduction?
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   Reduction

5. What does a dilation preserve?
   Show the full solution

   Angle measure, and therefore shape

6. A triangle with perimeter 20 is dilated with \( k = 3 \). Find the new perimeter and the ratio of areas.
   Show the full solution

   Perimeter scales by \( k \): \( 20 \times 3 = 60 \). Area scales by \( k^2 \): the ratio is \( 3^2 = 9 \), so the new area is nine times the old. Perimeter 60; area nine times larger

7. A dilation maps \( (6, 9) \) to \( (4, 6) \). Find \( k \) and the image of \( (3, 0) \).
   Show the full solution

   Comparing coordinates: \( \dfrac{4}{6} = \dfrac{2}{3} \) and \( \dfrac{6}{9} = \dfrac{2}{3} \). Both give the same factor, confirming a dilation about the origin with \( k = \dfrac{2}{3} \). Image of \( (3, 0) \): \( \left( \dfrac{2}{3}(3), \dfrac{2}{3}(0) \right) = (2, 0) \). \( k = \frac{2}{3} \); the image is \( (2, 0) \)

8. A square of side 5 is dilated so its area becomes 225. Find \( k \).
   Show the full solution

   The original area is \( 5^2 = 25 \). Area scales by \( k^2 \), so \( 25k^2 = 225 \), giving \( k^2 = 9 \) and \( k = 3 \), taking the positive root since a scale factor is positive. Check: the new side is \( 5 \times 3 = 15 \), and \( 15^2 = 225 \). Correct. The trap is computing \( \dfrac{225}{25} = 9 \) and calling that the scale factor. That ratio is \( k^2 \), not \( k \), and a square root is needed. \( k = 3 \)

9. Explain why a line not through the center of a dilation maps to a parallel line.
   Show the full solution

   Take a dilation about the origin with factor \( k \), and a line not through the origin. Pick two points on it, \( P(x_1, y_1) \) and \( Q(x_2, y_2) \). Their images are \( P'(kx_1, ky_1) \) and \( Q'(kx_2, ky_2) \). The slope of \( \overleftrightarrow{PQ} \) is \( \dfrac{y_2 - y_1}{x_2 - x_1} \). The slope of \( \overleftrightarrow{P'Q'} \) is \( \dfrac{ky_2 - ky_1}{kx_2 - kx_1} = \dfrac{k(y_2 - y_1)}{k(x_2 - x_1)} = \dfrac{y_2 - y_1}{x_2 - x_1} \), since the \( k \) cancels. Equal slopes, so the lines are parallel. They are distinct because the original line misses the origin and every point on it moved by a nonzero amount along a ray from the origin. The exception. If the line passes through the center, every point on it stays on the same ray from the center, so the image is the same line rather than a parallel one. The slope computation still gives the same slope; what changes is that the lines coincide instead of being distinct. The scale factor cancels in the slope calculation, so the image has the same slope; it is a different line unless the original passed through the center

10. A dilation about the origin with \( k = 3 \) is followed by a dilation about the origin with \( k = \frac{1}{2} \). Find the single dilation equivalent to the composition, and test it on a point.
    Show the full solution

    Work out the composition algebraically. A point \( (x, y) \) goes first to \( (3x, 3y) \), then to \( \left( \dfrac{1}{2}(3x), \dfrac{1}{2}(3y) \right) = \left( \dfrac{3x}{2}, \dfrac{3y}{2} \right) \). So the composition is the rule \( (x,y) \to \left( \dfrac{3}{2}x, \dfrac{3}{2}y \right) \), a single dilation about the origin with \( k = \dfrac{3}{2} \). The pattern. The composed scale factor is the product of the two: \( 3 \times \dfrac{1}{2} = \dfrac{3}{2} \). Dilations about the same center compose by multiplying scale factors, which is why the inverse of a factor \( k \) is \( \dfrac{1}{k} \). Test on a point. Take \( (4, 6) \). First dilation: \( (12, 18) \). Second: \( (6, 9) \). Directly by the composed rule: \( \left( \dfrac{3}{2}(4), \dfrac{3}{2}(6) \right) = (6, 9) \). They agree. Note on order. Unlike the compositions of lesson 4.5, these commute: applying \( \dfrac{1}{2} \) first gives \( (2, 3) \) and then \( 3 \) gives \( (6, 9) \), the same point. That is because multiplication of scale factors is commutative, and it holds only when the dilations share a center. A single dilation with \( k = \frac{3}{2} \), since scale factors about the same center multiply

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What symbol means "is similar to"?
   Show the full solution

   \( \sim \)

2. Two similar figures have sides in the ratio \( 2 : 5 \). What is the ratio of their perimeters?
   Show the full solution

   \( 2 : 5 \)

3. What is the ratio of their areas?
   Show the full solution

   Square the scale factor. \( 4 : 25 \)

4. Are all squares similar?
   Show the full solution

   All angles are right and all sides proportional. Yes

5. Are all rectangles similar?
   Show the full solution

   A 1 by 2 and a 1 by 5 have unequal side ratios. No

6. \( \triangle ABC \sim \triangle DEF \) with \( AB = 10 \), \( DE = 15 \), \( AC = 14 \). Find \( DF \).
   Show the full solution

   Scale factor: \( k = \dfrac{15}{10} = \dfrac{3}{2} \). \( \overline{AC} \) corresponds to \( \overline{DF} \), so \( DF = \dfrac{3}{2} \times 14 = 21 \). Check by proportion: \( \dfrac{10}{15} = \dfrac{14}{21} \), and cross multiplying gives \( 210 = 210 \). Correct. \( DF = 21 \)

7. Two similar triangles have areas 18 and 50. Find the ratio of their sides.
   Show the full solution

   The area ratio is \( \dfrac{18}{50} = \dfrac{9}{25} \). Areas are in the ratio \( k^2 \), so \( k^2 = \dfrac{9}{25} \) and \( k = \dfrac{3}{5} \), taking the positive root. So the sides are in the ratio \( 3 : 5 \). Check: \( \left( \dfrac{3}{5} \right)^2 = \dfrac{9}{25} = \dfrac{18}{50} \). Correct. \( 3 : 5 \)

8. Two similar polygons have perimeters 24 and 36, and the smaller has area 32. Find the larger area.
   Show the full solution

   Perimeters are in the ratio of the scale factor, so \( k = \dfrac{36}{24} = \dfrac{3}{2} \). Areas are in the ratio \( k^2 = \dfrac{9}{4} \). Larger area: \( \dfrac{9}{4} \times 32 = 72 \). Check: \( \dfrac{72}{32} = 2.25 = \left( \dfrac{3}{2} \right)^2 \). Correct. Note the two different scalings used in one problem: the perimeter gave \( k \) directly, and the area required \( k^2 \). 72

9. Explain why the area ratio is the square of the scale factor.
   Show the full solution

   A dilation multiplies every length by \( k \), and an area is measured in two dimensions. Take the simplest case, a rectangle with base \( b \) and height \( h \), so its area is \( bh \). After a dilation the base is \( kb \) and the height is \( kh \), so the new area is \( (kb)(kh) = k^2 bh \). Both factors picked up a \( k \), so the product picked up \( k^2 \). For a triangle the same happens: \( \dfrac{1}{2}(kb)(kh) = k^2 \cdot \dfrac{1}{2}bh \). For any region the argument works by approximating it with small rectangles, each of which scales by \( k^2 \), so the total does too. The practical consequence. Doubling the dimensions of a shape quadruples its area, which is why a pizza of twice the diameter feeds four people rather than two. And in three dimensions volume scales by \( k^3 \), which lesson 11.6 takes up. An area involves two length measurements, each scaled by \( k \), so their product scales by \( k^2 \)

10. A photograph 4 inches by 6 inches is enlarged so its area is 216 square inches. Find the new dimensions and the scale factor, and check the shape is preserved.
    Show the full solution

    Original area. \( 4 \times 6 = 24 \) square inches. Find the scale factor. The area ratio is \( \dfrac{216}{24} = 9 \), and this equals \( k^2 \), so \( k = 3 \). New dimensions. Each length scales by 3: \( 4 \times 3 = 12 \) and \( 6 \times 3 = 18 \) inches. Check the area. \( 12 \times 18 = 216 \). Correct. Check the shape. The original ratio of sides is \( \dfrac{4}{6} = \dfrac{2}{3} \). The new ratio is \( \dfrac{12}{18} = \dfrac{2}{3} \). Equal, so the rectangles are similar and the photograph is not distorted. Why the check matters. An enlargement that changed the side ratio would stretch the image. Preserving the ratio is what makes it a dilation rather than the shear or one-directional stretch of lesson 4.1. A commercial print size of 8 by 12 would also be similar, since \( \dfrac{8}{12} = \dfrac{2}{3} \), but 5 by 7 would not, since \( \dfrac{5}{7} \neq \dfrac{2}{3} \), which is why 4 by 6 photographs are cropped to fit 5 by 7 frames. 12 by 18 inches with \( k = 3 \); the side ratio stays \( \frac{2}{3} \), so the shape is preserved

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does AA require?
   Show the full solution

   Two pairs of congruent angles

2. Why is a third angle unnecessary?
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   The angle sum determines it

3. Two triangles have sides 3, 4, 5 and 6, 8, 10. Are they similar?
   Show the full solution

   All ratios equal \( \frac{1}{2} \). Yes, by SSS similarity

4. \( \triangle ABC \sim \triangle DEF \). Write the correct proportion relating \( AB \), \( DE \), \( BC \), \( EF \).
   Show the full solution

   \( \dfrac{AB}{DE} = \dfrac{BC}{EF} \)

5. Are all equilateral triangles similar?
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   All angles are \( 60^\circ \), so AA applies. Yes

6. Two triangles have sides 4, 6 and an included angle of \( 50^\circ \), and sides 10, 15 with an included angle of \( 50^\circ \). Are they similar?
   Show the full solution

   Check the side ratios: \( \dfrac{4}{10} = \dfrac{2}{5} \) and \( \dfrac{6}{15} = \dfrac{2}{5} \). Equal, so those two pairs are proportional. The included angles are both \( 50^\circ \), so they are congruent. Two proportional sides with congruent included angles is SAS similarity. Yes, by SAS similarity

7. In a triangle, a line parallel to one side cuts the other two. The small triangle has sides 5 and 7, and the corresponding sides of the whole are 15 and \( x \). Find \( x \).
   Show the full solution

   A line parallel to a side creates a triangle similar to the whole, by AA as in the worked example. The scale factor from small to large is \( \dfrac{15}{5} = 3 \). So \( x = 7 \times 3 = 21 \). Check by proportion: \( \dfrac{5}{15} = \dfrac{7}{21} \), and cross multiplying gives \( 105 = 105 \). Correct. \( x = 21 \)

8. Two triangles have sides 5, 7, 9 and 10, 14, 17. Are they similar?
   Show the full solution

   Test all three ratios, pairing the sides in increasing order. \( \dfrac{5}{10} = \dfrac{1}{2} \). \( \dfrac{7}{14} = \dfrac{1}{2} \). \( \dfrac{9}{17} \approx 0.529 \), which is not \( \dfrac{1}{2} \). The third ratio differs, so the sides are not proportional and SSS similarity does not apply. No. The trap is checking only two ratios. Two matching ratios prove nothing on their own, and the third side would have to be 18 for the triangles to be similar. Not similar; the third ratio fails

9. Explain why AA works for similarity but AAA fails for congruence.
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   Angles determine shape and say nothing about size. Knowing three pairs of congruent angles fixes the shape completely: the triangles have the same proportions and differ only by a uniform scaling. That is exactly what similarity asserts, so AA, which gives the third angle free, is a valid similarity criterion. Congruence demands size as well. Two equilateral triangles with sides of 1 and of 100 have all three angle pairs congruent and are clearly not congruent. No angle information can distinguish them, because angles are unchanged by a dilation. The general principle: a criterion must pin down everything the conclusion asserts. Similarity asserts shape, and angles determine shape. Congruence asserts shape and size, so at least one side must appear among the given parts. That is why every congruence criterion contains an S and no similarity criterion needs to. Angles determine shape but not size; similarity claims only shape, while congruence claims size too and therefore needs a side

10. In \( \triangle ABC \), \( \overline{DE} \parallel \overline{BC} \) with \( D \) on \( \overline{AB} \) and \( E \) on \( \overline{AC} \). Given \( AD = 6 \), \( DB = 4 \), \( AE = 9 \), find \( EC \), \( AC \), and the ratio of the areas of \( \triangle ADE \) and \( \triangle ABC \).
    Show the full solution

    Establish the similarity. Since \( \overline{DE} \parallel \overline{BC} \), corresponding angles give \( \angle ADE \cong \angle ABC \), and \( \angle A \) is shared. By AA, \( \triangle ADE \sim \triangle ABC \). Find \( AB \). \( AB = AD + DB = 6 + 4 = 10 \). Find the scale factor. From small to large, \( k = \dfrac{AB}{AD} = \dfrac{10}{6} = \dfrac{5}{3} \). Find \( AC \). \( \overline{AE} \) corresponds to \( \overline{AC} \), so \( AC = \dfrac{5}{3} \times 9 = 15 \). Find \( EC \). \( EC = AC - AE = 15 - 9 = 6 \). Check with the proportionality theorem of lesson 8.5. \( \dfrac{AD}{DB} = \dfrac{6}{4} = \dfrac{3}{2} \) and \( \dfrac{AE}{EC} = \dfrac{9}{6} = \dfrac{3}{2} \). Equal, as the theorem requires. Find the area ratio. Areas are in the ratio \( k^2 \). Going from \( \triangle ADE \) to \( \triangle ABC \) the factor is \( \dfrac{5}{3} \), so the area ratio is \( \left( \dfrac{3}{5} \right)^2 = \dfrac{9}{25} \) for small to large. So \( \triangle ADE \) has \( \dfrac{9}{25} \) of the area of \( \triangle ABC \), and the trapezoid \( DBCE \) has the remaining \( \dfrac{16}{25} \). A point worth noticing. The small triangle occupies 36 percent of the area while its sides are 60 percent as long. That gap is the squaring, and it is why a line drawn parallel to a side to cut off "half the triangle" must be placed at \( \dfrac{1}{\sqrt{2}} \approx 0.707 \) of the way, not halfway. \( EC = 6 \), \( AC = 15 \), and the areas are in the ratio \( 9 : 25 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the triangle proportionality theorem.
   Show the full solution

   A line parallel to one side of a triangle divides the other two proportionally

2. If \( \dfrac{AD}{DB} = \dfrac{3}{4} \) and \( AE = 9 \), find \( EC \).
   Show the full solution

   \( \dfrac{3}{4} = \dfrac{9}{EC} \), so \( 3 \cdot EC = 36 \). \( EC = 12 \)

3. What does the converse prove?
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   That the line is parallel to the third side

4. \( AD = 3 \), \( DB = 5 \), \( AE = 6 \), \( EC = 10 \). Is \( \overline{DE} \parallel \overline{BC} \)?
   Show the full solution

   \( \frac{3}{5} = \frac{6}{10} \). Yes

5. \( AD = 4 \), \( DB = 5 \), \( AE = 8 \), \( EC = 9 \). Is \( \overline{DE} \parallel \overline{BC} \)?
   Show the full solution

   \( \frac{4}{5} = 0.8 \) but \( \frac{8}{9} \approx 0.889 \). No

6. In \( \triangle ABC \) with \( \overline{DE} \parallel \overline{BC} \), \( AD = 2x \), \( DB = 12 \), \( AE = x + 3 \), \( EC = 9 \). Find \( x \).
   Show the full solution

   By the theorem, \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \): \( \dfrac{2x}{12} = \dfrac{x + 3}{9} \). Cross multiplying: \( 18x = 12(x + 3) = 12x + 36 \), so \( 6x = 36 \) and \( x = 6 \). Check: \( AD = 12 \), \( AE = 9 \). Then \( \dfrac{12}{12} = 1 \) and \( \dfrac{9}{9} = 1 \). Equal. \( x = 6 \)

7. Three parallel lines cut one transversal into segments of 6 and 9. They cut a second transversal so the first segment is 8. Find the second.
   Show the full solution

   By the three-parallel-lines theorem, the transversals are cut proportionally: \( \dfrac{6}{9} = \dfrac{8}{x} \). Cross multiplying: \( 6x = 72 \), so \( x = 12 \). Check: \( \dfrac{6}{9} = \dfrac{2}{3} \) and \( \dfrac{8}{12} = \dfrac{2}{3} \). Equal. This theorem is the reason ruled paper can be used to divide a segment into equal parts: evenly spaced parallel lines cut any transversal into equal pieces. 12

8. An angle bisector from \( A \) meets \( \overline{BC} \) at \( D \), with \( AB = 8 \), \( AC = 12 \), \( BD = 6 \). Find \( DC \).
   Show the full solution

   The angle bisector of a triangle divides the opposite side in the ratio of the adjacent sides: \( \dfrac{BD}{DC} = \dfrac{AB}{AC} \). Substituting: \( \dfrac{6}{DC} = \dfrac{8}{12} = \dfrac{2}{3} \). Cross multiplying: \( 2 \cdot DC = 18 \), so \( DC = 9 \). Check: \( \dfrac{6}{9} = \dfrac{2}{3} \) and \( \dfrac{8}{12} = \dfrac{2}{3} \). Equal. Note that \( BC = 6 + 9 = 15 \), and checking the triangle inequality: \( 8 + 12 = 20 \gt 15 \). Valid. \( DC = 9 \)

9. Explain why the piece-to-piece and whole-to-whole forms of the proportion are equivalent.
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   Start from the whole form, which comes directly from the similar triangles: \( \dfrac{AD}{AB} = \dfrac{AE}{AC} \). Substitute \( AB = AD + DB \) and \( AC = AE + EC \): \( \dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC} \). Cross multiply: \( AD(AE + EC) = AE(AD + DB) \), which expands to \( AD \cdot AE + AD \cdot EC = AE \cdot AD + AE \cdot DB \). The term \( AD \cdot AE \) appears on both sides and cancels, leaving \( AD \cdot EC = AE \cdot DB \), which is the cross-multiplied form of \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \). Every step is reversible, so the two forms imply each other. The practical point. A problem supplies whichever lengths it supplies, and knowing both forms means no conversion is needed. What must not happen is mixing them: \( \dfrac{AD}{DB} = \dfrac{AE}{AC} \) compares a piece to a piece on one side and a piece to a whole on the other, and it is false. Substituting the segment sums and canceling the common term converts one into the other, and every step reverses

10. In \( \triangle ABC \), \( \overline{DE} \parallel \overline{BC} \) with \( AD = x \), \( DB = x + 4 \), \( AE = x - 1 \), \( EC = x + 4 \). Find \( x \) and all four lengths.
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    Set up. By the theorem, \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \), that is \( \dfrac{x}{x + 4} = \dfrac{x - 1}{x + 4} \). Notice the denominators. They are identical, so the fractions are equal exactly when the numerators are: \( x = x - 1 \), which gives \( 0 = -1 \), a contradiction. Conclusion: no solution exists. There is no value of \( x \) making this configuration possible with \( \overline{DE} \parallel \overline{BC} \). What this means geometrically. The two lower pieces \( DB \) and \( EC \) are equal, both \( x + 4 \). For the proportion to hold, the two upper pieces would also have to be equal, so \( AD = AE \). The problem gives them as \( x \) and \( x - 1 \), which differ by 1 for every \( x \). The data is inconsistent with the parallelism. How to fix it. If \( AE \) were \( x \) instead of \( x - 1 \), the proportion would hold for every positive \( x \), and the figure would be an isosceles configuration with the parallel line cutting both sides at the same ratio. The lesson. A proportion that reduces to a false numerical statement means the given data is impossible, not that the algebra went wrong. Recognizing that is worth more than forcing an answer. No solution; equal denominators force equal numerators, and \( x \) cannot equal \( x - 1 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the geometric mean of 4 and 9.
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   \( \sqrt{36} \). 6

2. Find the geometric mean of 3 and 27.
   Show the full solution

   \( \sqrt{81} \). 9

3. The altitude to the hypotenuse divides it into 4 and 9. Find the altitude.
   Show the full solution

   \( \sqrt{4 \times 9} \). 6

4. How many triangles are similar in this configuration?
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   Three: the original and the two small ones

5. A leg is the geometric mean of which two lengths?
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   The whole hypotenuse and the segment adjacent to that leg

6. The hypotenuse is 20, divided into 4 and 16. Find the altitude and both legs.
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   Altitude: \( h = \sqrt{4 \times 16} = \sqrt{64} = 8 \). Leg adjacent to the 4: \( \sqrt{4 \times 20} = \sqrt{80} = 4\sqrt{5} \approx 8.94 \). Leg adjacent to the 16: \( \sqrt{16 \times 20} = \sqrt{320} = 8\sqrt{5} \approx 17.89 \). Check with the Pythagorean theorem: \( 80 + 320 = 400 = 20^2 \). Correct. Check a small triangle: \( 4^2 + 8^2 = 16 + 64 = 80 = (4\sqrt{5})^2 \). Correct. Altitude 8; legs \( 4\sqrt{5} \) and \( 8\sqrt{5} \)

7. A right triangle has legs 9 and 12. Find the altitude to the hypotenuse.
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   First find the hypotenuse: \( \sqrt{81 + 144} = \sqrt{225} = 15 \). Method one, by area. The area is \( \dfrac{1}{2}(9)(12) = 54 \), computed using the legs as base and height. Using the hypotenuse as the base, the area is \( \dfrac{1}{2}(15)(h) \). Setting them equal: \( \dfrac{15h}{2} = 54 \), so \( h = \dfrac{108}{15} = 7.2 \). Method two, by the segments. Each leg is the geometric mean of the hypotenuse and its adjacent segment, so \( 9^2 = 15p \) gives \( p = 5.4 \), and \( 12^2 = 15q \) gives \( q = 9.6 \). Check: \( 5.4 + 9.6 = 15 \). Correct. Then \( h = \sqrt{pq} = \sqrt{5.4 \times 9.6} = \sqrt{51.84} = 7.2 \). Agrees. 7.2

8. The altitude to the hypotenuse is 6 and one segment is 4. Find the other segment and the hypotenuse.
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   The altitude is the geometric mean of the segments: \( 6 = \sqrt{4q} \), so \( 36 = 4q \) and \( q = 9 \). The hypotenuse is the sum: \( 4 + 9 = 13 \). Check by finding the legs: \( \sqrt{4 \times 13} = \sqrt{52} = 2\sqrt{13} \) and \( \sqrt{9 \times 13} = \sqrt{117} = 3\sqrt{13} \). Then \( 52 + 117 = 169 = 13^2 \). Correct. Other segment 9, hypotenuse 13

9. Prove that the two small triangles are similar to the original.
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   Let \( \triangle ABC \) have a right angle at \( C \), and let \( \overline{CD} \) be the altitude to hypotenuse \( \overline{AB} \), with \( D \) on \( \overline{AB} \). \( \triangle ACD \) and \( \triangle ABC \). They share \( \angle A \), so that is one congruent pair by the reflexive property. \( \angle ADC \) is a right angle, since \( \overline{CD} \) is an altitude and therefore perpendicular to \( \overline{AB} \). And \( \angle ACB \) is a right angle by hypothesis. Both are right angles, so they are congruent by the right angle congruence theorem. Two pairs of congruent angles gives \( \triangle ACD \sim \triangle ABC \) by AA. \( \triangle CBD \) and \( \triangle ABC \). They share \( \angle B \), and \( \angle CDB \) and \( \angle ACB \) are both right angles. By AA, \( \triangle CBD \sim \triangle ABC \). The two small ones. Each is similar to \( \triangle ABC \), so by the transitivity of similarity they are similar to each other: \( \triangle ACD \sim \triangle CBD \). The correspondence worth noting. In the similarity \( \triangle ACD \sim \triangle ABC \), the vertex \( A \) matches \( A \), \( C \) matches \( B \), and \( D \) matches \( C \). Writing the letters in that order is what makes the resulting proportions come out correctly, and it is where most errors in this configuration occur. Proved by AA twice, then transitivity

10. A right triangle has hypotenuse 34 with one segment 16 units from the altitude's foot. Find every length in the figure and verify three ways.
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    Segments. One is 16, so the other is \( 34 - 16 = 18 \). Altitude. \( h = \sqrt{16 \times 18} = \sqrt{288} = 12\sqrt{2} \approx 16.97 \). Legs. Adjacent to the 16: \( \sqrt{16 \times 34} = \sqrt{544} = 4\sqrt{34} \approx 23.32 \). Adjacent to the 18: \( \sqrt{18 \times 34} = \sqrt{612} = 6\sqrt{17} \approx 24.74 \). Verification one, the whole triangle. \( 544 + 612 = 1156 \), and \( 34^2 = 1156 \). Correct. Verification two, the small triangle on the 16. \( 16^2 + (12\sqrt{2})^2 = 256 + 288 = 544 \), which is the square of that leg. Correct. Verification three, the small triangle on the 18. \( 18^2 + 288 = 324 + 288 = 612 \), the square of the other leg. Correct. A fourth check, by area. Using the legs: \( \dfrac{1}{2}\sqrt{544}\sqrt{612} = \dfrac{1}{2}\sqrt{332928} = \dfrac{1}{2}(577.0) \approx 288.5 \). Using the hypotenuse and altitude: \( \dfrac{1}{2}(34)(12\sqrt{2}) = 17 \times 16.97 \approx 288.5 \). Agrees. Sanity check on the shape. The two segments 16 and 18 are nearly equal, so the altitude falls near the middle of the hypotenuse and the triangle is close to isosceles. Correspondingly the two legs, about 23.3 and 24.7, are close together. Consistent. Segments 16 and 18, altitude \( 12\sqrt{2} \), legs \( 4\sqrt{34} \) and \( 6\sqrt{17} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Why are shadow triangles similar?
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   The sun's rays are parallel, so the angles of elevation are equal, and both objects are perpendicular to the ground

2. A 5 ft person casts a 2 ft shadow. A pole casts a 10 ft shadow. Find the pole's height.
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   \( \frac{5}{2} = \frac{h}{10} \). 25 ft

3. A scale is 1 : 50. A model is 6 cm long. Find the real length.
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   \( 6 \times 50 \). 300 cm, or 3 m

4. Two similar triangles have a scale factor of 4. A side of 7 corresponds to what?
   Show the full solution

   28

5. What principle makes mirror problems work?
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   The angle of incidence equals the angle of reflection

6. A 4 ft stick casts a 6 ft shadow while a building casts a 96 ft shadow. Find the building's height.
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   The triangles are similar by AA, so \( \dfrac{4}{6} = \dfrac{h}{96} \). Cross multiplying: \( 6h = 384 \), so \( h = 64 \). Check: the building's shadow is 16 times the stick's, so the building should be 16 times as tall: \( 4 \times 16 = 64 \). Consistent, and 64 feet is plausible for a building. 64 feet

7. A mirror lies 20 ft from a tower. A person with eyes 5.5 ft high stands 4 ft from the mirror and sees the top. Find the tower's height.
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   The triangles are similar by AA, using the equal incidence and reflection angles and the two right angles at the ground. \( \dfrac{5.5}{4} = \dfrac{h}{20} \). Cross multiplying: \( 4h = 110 \), so \( h = 27.5 \). Check: the tower is 5 times as far from the mirror as the person, so it should be 5 times the eye height: \( 5.5 \times 5 = 27.5 \). Consistent. 27.5 feet

8. A map scale is 1 inch to 25 miles. Two cities are 3.6 inches apart on the map. Find the real distance, and find the map distance for cities 140 miles apart.
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   First part. \( \dfrac{1}{25} = \dfrac{3.6}{d} \), so \( d = 25 \times 3.6 = 90 \) miles. Second part. \( \dfrac{1}{25} = \dfrac{m}{140} \), so \( 25m = 140 \) and \( m = 5.6 \) inches. Check both against the scale: 90 miles at 25 miles per inch is \( \dfrac{90}{25} = 3.6 \) inches. Correct. And 5.6 inches at 25 miles per inch is \( 5.6 \times 25 = 140 \) miles. Correct. 90 miles; 5.6 inches

9. Explain why an inverted proportion often produces a plausible-looking wrong answer, and how to catch it.
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   An inverted proportion multiplies by the reciprocal of the correct factor, so the error scales the answer rather than producing nonsense. In the worked example the correct tree height was 45 feet and the inverted version gave 20 feet. Both are possible tree heights, so the answer does not announce itself as wrong. The reliable catch is a ratio check performed before or after the algebra. Ask how many times larger one known quantity is, then confirm the answer is larger by the same factor. The tree's shadow is \( \dfrac{30}{4} = 7.5 \) times the person's, so the tree must be 7.5 times the person's height, which is 45 rather than 20. A second habit that prevents it entirely: label the proportion in words before substituting numbers. Writing \( \dfrac{\text{small height}}{\text{small shadow}} = \dfrac{\text{large height}}{\text{large shadow}} \) makes the correspondence explicit, and substituting into a labeled equation is much harder to get backward than assembling four numbers into two fractions. Inversion scales the answer by the reciprocal, which usually still looks reasonable; the fix is a ratio check or writing the proportion in words first

10. A model of a building uses a scale of 1 : 200. The model has a floor area of 0.5 square meters and a volume of 0.1 cubic meters. Find the real floor area and volume.
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    The scale factor. Lengths scale by \( k = 200 \). Area. Areas scale by \( k^2 = 200^2 = 40{,}000 \). Real floor area: \( 0.5 \times 40{,}000 = 20{,}000 \) square meters. Volume. Volumes scale by \( k^3 = 200^3 = 8{,}000{,}000 \). Real volume: \( 0.1 \times 8{,}000{,}000 = 800{,}000 \) cubic meters. Sanity check. A floor area of 20,000 square meters is about 2 hectares, large but plausible for a major building's total floor area. A volume of 800,000 cubic meters corresponds to a building roughly 100 by 100 meters in plan and 80 meters tall, which is consistent. The point of the problem. Using \( k \) rather than \( k^2 \) for the area would give \( 0.5 \times 200 = 100 \) square meters, which is the size of an apartment rather than a building. Using \( k \) for the volume would give 20 cubic meters, smaller than a single room. The errors are not subtle once the answer is checked against reality, which is why the sanity check is worth the ten seconds. The three scalings. Length by \( k \), area by \( k^2 \), volume by \( k^3 \). Lesson 11.6 develops this fully, including why it explains such things as why large animals have proportionally thicker legs. Floor area 20,000 square meters; volume 800,000 cubic meters

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A right triangle has legs 3 and 4. Find the hypotenuse.
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   5

2. A right triangle has legs 5 and 12. Find the hypotenuse.
   Show the full solution

   13

3. A right triangle has hypotenuse 17 and a leg 8. Find the other leg.
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   \( 289 - 64 = 225 \). 15

4. Classify the triangle with sides 9, 12, 15.
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   \( 81 + 144 = 225 = 15^2 \). Right

5. Which side is the hypotenuse?
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   The longest, opposite the right angle

6. Classify the triangle with sides 10, 12, 16.
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   Triangle inequality: \( 10 + 12 = 22 \gt 16 \). Valid. Longest is 16. \( 10^2 + 12^2 = 100 + 144 = 244 \), and \( 16^2 = 256 \). Since \( 244 \lt 256 \), the triangle is obtuse. Obtuse

7. Classify the triangle with sides 11, 13, 16.
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   Triangle inequality: \( 11 + 13 = 24 \gt 16 \). Valid. Longest is 16. \( 11^2 + 13^2 = 121 + 169 = 290 \), and \( 16^2 = 256 \). Since \( 290 \gt 256 \), the triangle is acute. Acute

8. A rectangle is 7 by 24. Find the length of its diagonal.
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   A diagonal of a rectangle is the hypotenuse of a right triangle whose legs are two sides, since a rectangle has right angles. \( d = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \). This is the 7-24-25 triple. 25

9. Explain why the theorem and its converse do different jobs.
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   The theorem takes a right triangle as given and produces a numerical relationship among its sides. Its job is computation: knowing two sides of a right triangle, find the third. The converse takes a numerical relationship as given and produces the conclusion that an angle is right. Its job is classification: knowing three lengths, decide what kind of triangle they form. The distinction shows up in how each appears in a proof. A line citing the theorem must already have a right angle available, either given or established. A line citing the converse must already have the squared relationship, and it concludes the right angle rather than assuming it. Mixing them up produces circular reasoning in one direction and an unjustified assumption in the other. A proof that a triangle is right cannot cite the theorem, because the theorem's hypothesis is exactly what is being proved. The theorem computes a side in a triangle already known to be right; the converse establishes that a triangle is right from its side lengths

10. A ladder 25 feet long leans against a wall with its base 7 feet from the wall. Find how far up the wall it reaches. If the base slides out to 15 feet, how far does the top slide down?
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    First position. The ladder is the hypotenuse, the distance from the wall is one leg, and the height reached is the other. \( 7^2 + h^2 = 25^2 \), so \( h^2 = 625 - 49 = 576 \) and \( h = 24 \) feet. Second position. The ladder's length is unchanged at 25. \( 15^2 + h^2 = 25^2 \), so \( h^2 = 625 - 225 = 400 \) and \( h = 20 \) feet. The slide. The top moved from 24 feet to 20 feet, a drop of 4 feet. Something worth noticing. The base moved out by \( 15 - 7 = 8 \) feet and the top moved down by only 4 feet. The two do not change by equal amounts, because the relationship is quadratic rather than linear. Near the vertical position a small outward slide costs little height; near the horizontal it costs a great deal, which is the physical reason ladders become unstable as they flatten. Check both. \( 7^2 + 24^2 = 49 + 576 = 625 \) and \( 15^2 + 20^2 = 225 + 400 = 625 \). Both equal \( 25^2 \), so the ladder length is consistent. Both are recognizable triples, 7-24-25 and 3-4-5 scaled by 5. Reaches 24 feet; slides down 4 feet to 20 feet

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A \( 45 \)-\( 45 \)-\( 90 \) triangle has legs of 6. Find the hypotenuse.
   Show the full solution

   \( 6\sqrt{2} \)

2. A \( 30 \)-\( 60 \)-\( 90 \) triangle has a short leg of 5. Find the hypotenuse.
   Show the full solution

   Twice the short leg. 10

3. Find the long leg in that triangle.
   Show the full solution

   \( 5\sqrt{3} \)

4. Which leg is opposite the \( 30^\circ \) angle?
   Show the full solution

   The short leg

5. A square has side 8. Find its diagonal.
   Show the full solution

   The diagonal is the hypotenuse of a \( 45 \)-\( 45 \)-\( 90 \) triangle. \( 8\sqrt{2} \)

6. A \( 45 \)-\( 45 \)-\( 90 \) triangle has hypotenuse \( 12\sqrt{2} \). Find the legs.
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   The hypotenuse is \( x\sqrt{2} \), so \( x\sqrt{2} = 12\sqrt{2} \) and \( x = 12 \). Both legs measure 12. Check: \( 144 + 144 = 288 \) and \( (12\sqrt{2})^2 = 144 \times 2 = 288 \). Correct. 12 each

7. A \( 30 \)-\( 60 \)-\( 90 \) triangle has a long leg of \( 9\sqrt{3} \). Find the other two sides.
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   The long leg is \( x\sqrt{3} \), so \( x\sqrt{3} = 9\sqrt{3} \) and \( x = 9 \). Short leg: 9. Hypotenuse: \( 2x = 18 \). Check: \( 81 + 243 = 324 = 18^2 \). Correct. Ordering: \( 9 \lt 15.59 \lt 18 \). Consistent. Short leg 9, hypotenuse 18

8. An equilateral triangle has side 12. Find its height and area.
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   The altitude splits it into two \( 30 \)-\( 60 \)-\( 90 \) triangles with hypotenuse 12 and short leg \( \dfrac{12}{2} = 6 \). The height is the long leg: \( 6\sqrt{3} \approx 10.39 \). Area: \( \dfrac{1}{2}(12)(6\sqrt{3}) = 36\sqrt{3} \approx 62.35 \). Check the height against the Pythagorean theorem: \( 6^2 + (6\sqrt{3})^2 = 36 + 108 = 144 = 12^2 \). Correct. Sanity check: the height of about 10.39 is less than the side of 12, as it must be since the side is a hypotenuse. Height \( 6\sqrt{3} \approx 10.39 \), area \( 36\sqrt{3} \approx 62.35 \)

9. Explain why the ratios of these triangles are the same regardless of size.
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   Any two triangles with the same three angles are similar, by the AA criterion of lesson 8.4. All \( 45 \)-\( 45 \)-\( 90 \) triangles have the same three angles, so they are all similar to one another, and similar triangles have proportional sides. Proportional sides mean that dividing any side by any other gives the same value in every such triangle. So the ratio leg to hypotenuse is \( \dfrac{1}{\sqrt{2}} \) in a tiny one and in an enormous one alike. The same holds for the \( 30 \)-\( 60 \)-\( 90 \) family. That is exactly why memorizing the ratios is useful: they are properties of the shape rather than of any particular triangle, so knowing one side determines all three. The same idea makes trigonometry possible. Lesson 9.3 defines sine, cosine and tangent as ratios in a right triangle, and they are well defined for precisely this reason: all right triangles with a given acute angle are similar, so the ratio depends only on the angle. The special triangles are the two cases where those ratios come out as exact radicals rather than decimals. All triangles with the same angles are similar, so their side ratios are equal regardless of size

10. A regular hexagon has side 10. Find its area using special right triangles.
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    Decompose the hexagon. Drawing segments from the center to each vertex cuts a regular hexagon into six triangles. Each has a central angle of \( \dfrac{360}{6} = 60^\circ \), and its other two sides are radii and therefore congruent, so each triangle is isosceles with a \( 60^\circ \) vertex angle. Its base angles are each \( \dfrac{180 - 60}{2} = 60^\circ \), so every triangle is equilateral with side 10. Find the area of one triangle. Its altitude splits it into two \( 30 \)-\( 60 \)-\( 90 \) triangles with hypotenuse 10 and short leg 5, so the height is \( 5\sqrt{3} \). Area: \( \dfrac{1}{2}(10)(5\sqrt{3}) = 25\sqrt{3} \approx 43.30 \). Total area. Six such triangles: \( 6 \times 25\sqrt{3} = 150\sqrt{3} \approx 259.81 \). Check with the apothem formula of lesson 11.2. The apothem is the height of one triangle, \( 5\sqrt{3} \), and the perimeter is \( 6 \times 10 = 60 \). The area of a regular polygon is \( \dfrac{1}{2} \times \text{apothem} \times \text{perimeter} = \dfrac{1}{2}(5\sqrt{3})(60) = 150\sqrt{3} \). Agrees. Sanity check. The hexagon fits inside a circle of radius 10, whose area is \( 100\pi \approx 314.16 \). The hexagon's 259.81 is less, as it must be, and it is about 83 percent of the circle, which is right for a hexagon inscribed in its circumcircle. \( 150\sqrt{3} \approx 259.81 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use the 5-12-13 triangle with \( \angle A \) opposite the 5.

1. Find \( \sin A \).
   Show the full solution

   \( \frac{5}{13} \)

2. Find \( \cos A \).
   Show the full solution

   \( \frac{12}{13} \)

3. Find \( \tan A \).
   Show the full solution

   \( \frac{5}{12} \)

4. Find \( \sin B \), where \( \angle B \) is the other acute angle.
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   Opposite \( \angle B \) is 12. \( \frac{12}{13} \)

5. Can \( \cos\theta = 1.2 \) for an acute angle?
   Show the full solution

   The adjacent leg is shorter than the hypotenuse. No

6. A right triangle has legs 8 and 15. Find all three ratios for the angle opposite the 8.
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   Hypotenuse: \( \sqrt{64 + 225} = \sqrt{289} = 17 \). Relative to the angle opposite the 8: opposite is 8, adjacent is 15, hypotenuse is 17. \( \sin = \dfrac{8}{17} \approx 0.471 \). \( \cos = \dfrac{15}{17} \approx 0.882 \). \( \tan = \dfrac{8}{15} \approx 0.533 \). Check: sine and cosine are both between 0 and 1. Correct. \( \frac{8}{17} \), \( \frac{15}{17} \), \( \frac{8}{15} \)

7. Find the exact values of \( \sin 45^\circ \), \( \cos 45^\circ \) and \( \tan 45^\circ \).
   Show the full solution

   Use the \( 45 \)-\( 45 \)-\( 90 \) triangle with legs 1 and hypotenuse \( \sqrt{2} \). \( \sin 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} \approx 0.707 \). \( \cos 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} \approx 0.707 \). \( \tan 45^\circ = \dfrac{1}{1} = 1 \). Sine and cosine are equal because the triangle is isosceles, so opposite and adjacent are the same length. \( \frac{\sqrt{2}}{2} \), \( \frac{\sqrt{2}}{2} \), and 1

8. Find the exact values of \( \sin 30^\circ \), \( \cos 30^\circ \) and \( \tan 30^\circ \).
   Show the full solution

   Use the \( 30 \)-\( 60 \)-\( 90 \) triangle with short leg 1, long leg \( \sqrt{3} \), hypotenuse 2. Relative to the \( 30^\circ \) angle, the opposite is 1 and the adjacent is \( \sqrt{3} \). \( \sin 30^\circ = \dfrac{1}{2} = 0.5 \). \( \cos 30^\circ = \dfrac{\sqrt{3}}{2} \approx 0.866 \). \( \tan 30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} \approx 0.577 \). Check: \( \sin 30^\circ = 0.5 \) exactly, which is worth remembering as the one clean value. \( \frac{1}{2} \), \( \frac{\sqrt{3}}{2} \), \( \frac{\sqrt{3}}{3} \)

9. Explain why the trigonometric ratios depend only on the angle and not on the triangle.
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   Take two right triangles that both contain an acute angle of measure \( \theta \). Each has a right angle and an angle of \( \theta \), so two pairs of angles are congruent and the triangles are similar by AA. Similar triangles have proportional corresponding sides, meaning there is a scale factor \( k \) such that every side of the second is \( k \) times the corresponding side of the first. Now compute the sine in the second triangle: \( \dfrac{k \cdot \text{opposite}}{k \cdot \text{hypotenuse}} = \dfrac{\text{opposite}}{\text{hypotenuse}} \), since the \( k \) cancels. That is the sine in the first triangle. The same cancellation happens for cosine and tangent. Why this matters. It is what makes a table or a calculator button possible. If the ratio depended on which triangle was drawn, there would be no such thing as "the sine of \( 35^\circ \)"; there would only be the sine of a particular triangle. Because all such triangles give the same answer, the value can be looked up once and used everywhere. All right triangles with a given acute angle are similar by AA, so the scale factor cancels out of every ratio

10. In a right triangle, \( \sin\theta = \dfrac{7}{25} \). Find \( \cos\theta \) and \( \tan\theta \) exactly, without finding \( \theta \).
    Show the full solution

    Build a triangle from the given ratio. Since \( \sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{7}{25} \), take a right triangle with opposite side 7 and hypotenuse 25. Any triangle with this ratio will do, since the answers depend only on \( \theta \). Find the adjacent side. By the Pythagorean theorem, \( 7^2 + a^2 = 25^2 \), so \( a^2 = 625 - 49 = 576 \) and \( a = 24 \). This is the 7-24-25 triple. Write the other ratios. \( \cos\theta = \dfrac{24}{25} = 0.96 \). \( \tan\theta = \dfrac{7}{24} \approx 0.292 \). Check with the Pythagorean identity. \( \sin^2\theta + \cos^2\theta = \dfrac{49}{625} + \dfrac{576}{625} = \dfrac{625}{625} = 1 \). Correct, and this identity is just the Pythagorean theorem divided through by \( c^2 \). Second check. \( \dfrac{\sin\theta}{\cos\theta} = \dfrac{7/25}{24/25} = \dfrac{7}{24} = \tan\theta \). Correct, and that relationship holds for every angle. Sanity check on size. A sine of 0.28 is small, so \( \theta \) is a small angle, near \( 16^\circ \). Correspondingly the cosine should be near 1, and 0.96 is. The tangent should be close to the sine for a small angle, and 0.292 is close to 0.28. All consistent. \( \cos\theta = \frac{24}{25} \), \( \tan\theta = \frac{7}{24} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Round to two decimal places.

1. Which ratio relates the opposite side and the hypotenuse?
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   Sine

2. Which ratio relates the two legs?
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   Tangent

3. Hypotenuse 10, angle \( 30^\circ \). Find the opposite side.
   Show the full solution

   \( 10 \sin 30^\circ = 10(0.5) \). 5

4. Hypotenuse 10, angle \( 60^\circ \). Find the adjacent side.
   Show the full solution

   \( 10 \cos 60^\circ = 10(0.5) \). 5

5. What mode must the calculator be in?
   Show the full solution

   Degree

6. An angle of \( 52^\circ \) has adjacent leg 14. Find the hypotenuse.
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   Adjacent and hypotenuse means cosine, with the unknown in the denominator: \( \cos 52^\circ = \dfrac{14}{h} \), so \( h = \dfrac{14}{\cos 52^\circ} \). With \( \cos 52^\circ \approx 0.6157 \): \( h \approx \dfrac{14}{0.6157} \approx 22.74 \). Check: the hypotenuse exceeds the leg. Correct. About 22.74

7. An angle of \( 67^\circ \) has opposite leg 25. Find the adjacent leg.
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   Two legs means tangent, with the unknown in the denominator: \( \tan 67^\circ = \dfrac{25}{a} \), so \( a = \dfrac{25}{\tan 67^\circ} \). With \( \tan 67^\circ \approx 2.3559 \): \( a \approx \dfrac{25}{2.3559} \approx 10.61 \). Check: \( 67^\circ \) exceeds \( 45^\circ \), so the opposite leg should be longer than the adjacent, and \( 25 \gt 10.61 \). Consistent. About 10.61

8. Solve the right triangle with an acute angle of \( 42^\circ \) and hypotenuse 18.
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   Third angle: \( 90 - 42 = 48^\circ \). Opposite leg: \( 18 \sin 42^\circ \approx 18 \times 0.6691 \approx 12.04 \). Adjacent leg: \( 18 \cos 42^\circ \approx 18 \times 0.7431 \approx 13.38 \). Check with the Pythagorean theorem: \( 12.04^2 + 13.38^2 \approx 145.0 + 179.0 = 324.0 \), and \( 18^2 = 324 \). Agrees. Ordering: \( 12.04 \lt 13.38 \lt 18 \), matching angles \( 42^\circ \), \( 48^\circ \), \( 90^\circ \). Consistent. Angles \( 42^\circ \), \( 48^\circ \), \( 90^\circ \); sides about 12.04, 13.38 and 18

9. Explain how to decide whether to multiply or divide.
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   Write the equation first and let the algebra decide, rather than guessing. If the unknown is in the numerator, as in \( \sin\theta = \dfrac{x}{c} \), multiplying both sides by \( c \) gives \( x = c \sin\theta \). Multiply. If the unknown is in the denominator, as in \( \sin\theta = \dfrac{a}{x} \), cross multiplying gives \( x \sin\theta = a \), so \( x = \dfrac{a}{\sin\theta} \). Divide. A useful rule of thumb: the unknown is in the denominator exactly when the unknown is the larger of the two sides in the ratio, since the denominator of sine or cosine is the hypotenuse. The check that catches the error. Sine and cosine of an acute angle are always less than 1. Multiplying by such a number makes a length smaller; dividing makes it larger. So if the answer should be larger than the given side, divide; if smaller, multiply. A hypotenuse computed as shorter than a leg means the operation was reversed. Write the equation and solve it algebraically; the check is that sine and cosine are less than 1, so multiplying shrinks and dividing enlarges

10. A ramp rises at \( 8^\circ \) and must reach a loading dock 4 feet high. Find the ramp's length and its horizontal extent, and comment on the design.
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    The ramp's length. The height 4 is opposite the \( 8^\circ \) angle and the ramp is the hypotenuse, so use sine with the unknown in the denominator: \( \sin 8^\circ = \dfrac{4}{L} \), so \( L = \dfrac{4}{\sin 8^\circ} \). With \( \sin 8^\circ \approx 0.1392 \): \( L \approx \dfrac{4}{0.1392} \approx 28.74 \) feet. The horizontal extent. The horizontal distance is adjacent to the \( 8^\circ \) angle, and the height is opposite, so use tangent: \( \tan 8^\circ = \dfrac{4}{d} \), so \( d = \dfrac{4}{\tan 8^\circ} \). With \( \tan 8^\circ \approx 0.1405 \): \( d \approx \dfrac{4}{0.1405} \approx 28.46 \) feet. Check with the Pythagorean theorem. \( 4^2 + 28.46^2 \approx 16 + 810.0 = 826.0 \), and \( 28.74^2 \approx 825.9 \). Agrees to rounding. Comment on the design. The ramp is nearly 29 feet long to gain 4 feet of height, and its horizontal run is almost the same as its length, since a \( 8^\circ \) angle is shallow. That shallowness is deliberate: accessibility guidelines commonly require a slope no steeper than 1 in 12, which is about \( 4.8^\circ \). At \( 8^\circ \) the slope is roughly 1 in 7, steeper than that standard, so a compliant ramp would need to be longer still, about 48 feet. The lesson about small angles. For a small angle the sine and tangent are nearly equal, 0.1392 against 0.1405 here, so the ramp length and the horizontal run come out within a foot of each other. That approximation fails badly for large angles and is worth knowing only as a sanity check, not as a shortcut. Ramp about 28.74 feet long with a horizontal run of about 28.46 feet; steeper than a typical 1 in 12 accessibility slope

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Round angles to two decimal places.

1. What does \( \sin^{-1} \) mean?
   Show the full solution

   The inverse sine function, not the reciprocal

2. Find \( \sin^{-1}(0.5) \).
   Show the full solution

   \( 30^\circ \)

3. Find \( \tan^{-1}(1) \).
   Show the full solution

   \( 45^\circ \)

4. Opposite 5, hypotenuse 13. Which function do you use?
   Show the full solution

   Inverse sine

5. Can \( \cos^{-1}(1.3) \) be computed?
   Show the full solution

   Cosine never exceeds 1 for an acute angle. No

6. A right triangle has adjacent leg 12 and hypotenuse 20. Find the angle.
   Show the full solution

   Adjacent and hypotenuse means cosine: \( \cos\theta = \dfrac{12}{20} = 0.6 \), so \( \theta = \cos^{-1}(0.6) \approx 53.13^\circ \). Check: \( \cos 53.13^\circ \approx 0.600 \). Correct. This is the 3-4-5 triangle scaled by 4. About \( 53.13^\circ \)

7. A right triangle has legs 9 and 5. Find both acute angles.
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   Angle opposite the 5: \( \tan\theta = \dfrac{5}{9} \approx 0.5556 \), so \( \theta = \tan^{-1}(0.5556) \approx 29.05^\circ \). Angle opposite the 9: \( 90 - 29.05 = 60.95^\circ \). Check independently: \( \tan^{-1}\left( \dfrac{9}{5} \right) = \tan^{-1}(1.8) \approx 60.95^\circ \). Agrees. Sum check: \( 29.05 + 60.95 = 90.00 \). Correct. About \( 29.05^\circ \) and \( 60.95^\circ \)

8. Solve completely the right triangle with hypotenuse 17 and one leg 8.
   Show the full solution

   Third side: \( \sqrt{289 - 64} = \sqrt{225} = 15 \). This is the 8-15-17 triple. Angle opposite the 8: \( \sin\theta = \dfrac{8}{17} \approx 0.4706 \), so \( \theta = \sin^{-1}(0.4706) \approx 28.07^\circ \). Angle opposite the 15: \( 90 - 28.07 = 61.93^\circ \). Check: \( \sin 61.93^\circ \approx 0.882 = \dfrac{15}{17} \). Correct. Ordering: \( 8 \lt 15 \lt 17 \) matching \( 28.07^\circ \), \( 61.93^\circ \), \( 90^\circ \). Consistent. Sides 8, 15, 17; angles about \( 28.07^\circ \), \( 61.93^\circ \), \( 90^\circ \)

9. Explain why an inverse sine input must be between 0 and 1 but an inverse tangent input need not be.
   Show the full solution

   Sine is the opposite leg divided by the hypotenuse. In any right triangle the hypotenuse is the longest side, proved in lesson 6.6, so the numerator is always smaller than the denominator and the ratio is strictly between 0 and 1. An inverse sine is asking which angle produces a given ratio, so an input outside that range corresponds to no triangle and no angle. The same argument applies to cosine, since the adjacent leg is also shorter than the hypotenuse. Tangent divides one leg by the other, and neither is required to be larger. For an angle under \( 45^\circ \) the opposite leg is shorter and the tangent is under 1; for an angle over \( 45^\circ \) it is over 1; at exactly \( 45^\circ \) it is 1. As the angle approaches \( 90^\circ \) the opposite leg grows without bound relative to the adjacent, so the tangent takes arbitrarily large values. Every positive number is the tangent of some acute angle. The practical use. An inverse sine or cosine error on a calculator is a reliable signal that a ratio was formed upside down, which is worth treating as diagnostic rather than as a nuisance. Sine and cosine divide a leg by the longer hypotenuse, so they are always under 1; tangent divides one leg by the other and either may be longer

10. A right triangle has legs \( a \) and \( 2a \). Find both acute angles, and explain why the answer does not depend on \( a \).
    Show the full solution

    Find the angles. Take the angle opposite the shorter leg \( a \). Its adjacent leg is \( 2a \), so \( \tan\theta = \dfrac{a}{2a} = \dfrac{1}{2} = 0.5 \), since the \( a \) cancels. \( \theta = \tan^{-1}(0.5) \approx 26.57^\circ \). The other acute angle is \( 90 - 26.57 = 63.43^\circ \). Why \( a \) does not matter. The variable canceled in the ratio, which is the whole point of lesson 9.3: the trigonometric ratios depend only on the angle, not on the size of the triangle. Every right triangle whose legs are in the ratio \( 1 : 2 \) is similar to every other, by SAS similarity with the included right angles, so they all have the same angles. Check with a concrete case. Take \( a = 3 \), so the legs are 3 and 6 and the hypotenuse is \( \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \approx 6.708 \). \( \sin\theta = \dfrac{3}{6.708} \approx 0.4472 \), and \( \sin^{-1}(0.4472) \approx 26.57^\circ \). Agrees. Take \( a = 100 \): legs 100 and 200, hypotenuse \( 100\sqrt{5} \approx 223.6 \), and \( \sin\theta = \dfrac{100}{223.6} \approx 0.4472 \), the same. Confirmed. Sum check. \( 26.57 + 63.43 = 90.00 \). Correct. About \( 26.57^\circ \) and \( 63.43^\circ \); the \( a \) cancels because the ratios depend only on shape

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. From where is an angle of elevation measured?
   Show the full solution

   From the horizontal

2. Is the angle of depression measured from the vertical?
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   No, from the horizontal

3. How do the angle of elevation and depression between two points compare?
   Show the full solution

   They are equal

4. A tree casts an angle of elevation of \( 45^\circ \) from 20 m away. Find its height, ignoring eye level.
   Show the full solution

   \( 20 \tan 45^\circ = 20(1) \). 20 m

5. What extra step is needed when the observer's eye height is given?
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   Add the eye height to the computed height above eye level

6. From 120 m away, the angle of elevation to a building's top is \( 28^\circ \). The observer's eyes are 1.5 m high. Find the building's height.
   Show the full solution

   Above eye level: \( 120 \tan 28^\circ \approx 120 \times 0.5317 \approx 63.80 \) m. Adding the eye height: \( 63.80 + 1.5 = 65.30 \) m. Check: the angle is under \( 45^\circ \), so the height above eye level should be less than the horizontal distance, and \( 63.80 \lt 120 \). Consistent. About 65.30 m

7. From a lighthouse 45 m tall, the angle of depression to a ship is \( 18^\circ \). Find the ship's distance from the base.
   Show the full solution

   The angle of depression equals the angle of elevation from the ship, so the \( 18^\circ \) angle sits at the ship with opposite side 45 and adjacent side \( d \). \( \tan 18^\circ = \dfrac{45}{d} \), so \( d = \dfrac{45}{\tan 18^\circ} \). With \( \tan 18^\circ \approx 0.3249 \): \( d \approx \dfrac{45}{0.3249} \approx 138.5 \) m. Check: a shallow angle means a distant ship, and 138.5 m is about three times the lighthouse height. Consistent. About 138.5 m

8. A kite string 80 m long makes an angle of \( 52^\circ \) with the ground. Find the kite's height above the ground, assuming the string is held at ground level.
   Show the full solution

   The string is the hypotenuse and the height is opposite the \( 52^\circ \) angle, so use sine: \( h = 80 \sin 52^\circ \approx 80 \times 0.7880 \approx 63.04 \) m. Check: the height must be less than the string length, and \( 63.04 \lt 80 \). Correct. A second check: the horizontal distance is \( 80 \cos 52^\circ \approx 49.25 \) m, and \( 63.04^2 + 49.25^2 \approx 3974 + 2426 = 6400 = 80^2 \). Agrees. About 63.04 m

9. Explain why drawing the diagram first prevents most errors in these problems.
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   Three specific errors are prevented by the drawing, and all three are hard to catch afterward. Measuring from the wrong reference. An angle of elevation is from the horizontal, and a student who imagines it from the vertical will use the complement, producing an answer that is wrong but not obviously so. Drawing the horizontal explicitly makes the reference visible. Mislabeling opposite and adjacent. Which leg is opposite the angle depends on where the angle sits, and in a word problem the angle may be at the top or at the bottom of the figure. A labeled diagram settles it before a ratio is chosen. Forgetting the observer's height. On a drawing, the eye-level line is visibly above the ground, and the gap between them is an obvious separate quantity. In an unillustrated calculation it is invisible. A fourth benefit. The diagram supports the sanity check. Seeing that the angle is shallow makes it obvious that the horizontal distance should exceed the height, so an answer with those reversed is caught immediately. It fixes the reference for the angle, settles which leg is opposite, makes the observer's height visible, and supports the plausibility check

10. From the top of a building, the angle of depression to the near edge of a road is \( 62^\circ \) and to the far edge is \( 41^\circ \). The building is 70 m tall. Find the road's width.
    Show the full solution

    Set up. Both angles are measured from the same point at the top of the building, down to two points on the ground at different distances. Each gives a right triangle sharing the vertical side of 70 m. The angles of depression equal the angles of elevation at the two edges, so each sits at ground level in its triangle. Near edge. The \( 62^\circ \) angle has opposite side 70 and adjacent side \( d_1 \): \( \tan 62^\circ = \dfrac{70}{d_1} \), so \( d_1 = \dfrac{70}{\tan 62^\circ} \approx \dfrac{70}{1.8807} \approx 37.22 \) m. Far edge. The \( 41^\circ \) angle gives \( d_2 = \dfrac{70}{\tan 41^\circ} \approx \dfrac{70}{0.8693} \approx 80.52 \) m. The road's width. The difference of the two distances: \( 80.52 - 37.22 = 43.30 \) m. Check the ordering. A steeper angle of depression corresponds to a nearer point, and \( 62^\circ \gt 41^\circ \) went with 37.22 m against 80.52 m. Consistent. Had the subtraction been done the other way the answer would be negative, which is an immediate signal. Sanity check. A road 43 m wide is broad but plausible for a major highway with several lanes and a median. And both distances are reasonable for viewing from a 70 m building, roughly a twenty-story tower. Check with the angles. Recomputing: \( \tan^{-1}\left( \dfrac{70}{37.22} \right) \approx 62.0^\circ \) and \( \tan^{-1}\left( \dfrac{70}{80.52} \right) \approx 41.0^\circ \). Both recover the given angles. About 43.30 m

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is \( \cos(90^\circ - \theta) \) equal to?
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   \( \sin\theta \)

2. \( \sin 25^\circ \) equals which cosine?
   Show the full solution

   \( \cos 65^\circ \)

3. \( \cos 18^\circ \) equals which sine?
   Show the full solution

   \( \sin 72^\circ \)

4. At what angle does \( \sin\theta = \cos\theta \)?
   Show the full solution

   It is its own complement. \( 45^\circ \)

5. Why are the two acute angles of a right triangle complementary?
   Show the full solution

   The three angles sum to \( 180^\circ \) and one is \( 90^\circ \)

6. Solve \( \sin(2x + 10^\circ) = \cos(x + 20^\circ) \).
   Show the full solution

   The relationship holds when the two angles are complementary: \( (2x + 10) + (x + 20) = 90 \), so \( 3x + 30 = 90 \), giving \( 3x = 60 \) and \( x = 20^\circ \). Check: the angles are \( 2(20) + 10 = 50^\circ \) and \( 20 + 20 = 40^\circ \), summing to \( 90^\circ \). And \( \sin 50^\circ \approx 0.766 = \cos 40^\circ \). Correct. \( x = 20^\circ \)

7. Given \( \sin 35^\circ \approx 0.5736 \), find \( \cos 55^\circ \) without a calculator.
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   Since \( 35 + 55 = 90 \), the angles are complementary, so \( \cos 55^\circ = \sin 35^\circ \approx 0.5736 \). No computation was needed beyond recognizing the complement. About 0.5736

8. In a right triangle, \( \sin A = \dfrac{3}{5} \). Find \( \cos B \), where \( \angle B \) is the other acute angle.
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   The two acute angles are complementary, so \( B = 90^\circ - A \). Therefore \( \cos B = \cos(90^\circ - A) = \sin A = \dfrac{3}{5} \). Confirming with the triangle: if \( \sin A = \dfrac{3}{5} \), take the opposite side 3 and hypotenuse 5, giving adjacent side 4. Relative to \( \angle B \), the side of 3 is now adjacent, so \( \cos B = \dfrac{3}{5} \). Agrees. \( \frac{3}{5} \)

9. Explain why cosine is named "cosine."
   Show the full solution

   The name is a contraction of "complement's sine," and it records exactly the relationship of this lesson: the cosine of an angle is the sine of its complement. The historical route is through Latin. Early trigonometric tables listed the sine of an angle and, alongside it, the sine of the complementary angle, written as sinus complementi. That phrase contracted to cosinus and then to cosine. The same pattern names the other co-functions. Cotangent is the tangent of the complement, which is why \( \tan\theta \) and \( \cot\theta \) are reciprocals of each other when the angles are complementary. Cosecant relates to secant the same way. Why this is worth knowing. The name is a mnemonic for the identity rather than an arbitrary label, so remembering what cosine means gives the relationship free. And it explains why the sine and cosine graphs in later courses are the same curve shifted by \( 90^\circ \): they are the same function evaluated at complementary angles. It contracts "sine of the complement," which is precisely the relationship \( \cos\theta = \sin(90^\circ - \theta) \)

10. Prove that \( \sin^2\theta + \cos^2\theta = 1 \) for any acute angle, and use it to find \( \cos\theta \) when \( \sin\theta = 0.28 \).
    Show the full solution

    The proof. Take a right triangle with the angle \( \theta \), opposite side \( a \), adjacent side \( b \), and hypotenuse \( c \). By definition, \( \sin\theta = \dfrac{a}{c} \) and \( \cos\theta = \dfrac{b}{c} \). Squaring and adding: \[ \sin^2\theta + \cos^2\theta = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2 + b^2}{c^2} \] By the Pythagorean theorem, \( a^2 + b^2 = c^2 \), so the fraction is \( \dfrac{c^2}{c^2} = 1 \). So the identity is the Pythagorean theorem with every term divided by \( c^2 \), which is why it is called the Pythagorean identity. Using it. Given \( \sin\theta = 0.28 \): \( \cos^2\theta = 1 - 0.28^2 = 1 - 0.0784 = 0.9216 \), so \( \cos\theta = \sqrt{0.9216} = 0.96 \), taking the positive root since \( \theta \) is acute and cosine is positive there. Check. \( \sin^{-1}(0.28) \approx 16.26^\circ \), and \( \cos 16.26^\circ \approx 0.960 \). Correct. A second check by triangle. A sine of 0.28 corresponds to a 7-24-25 triangle scaled, since \( \dfrac{7}{25} = 0.28 \). Then \( \cos\theta = \dfrac{24}{25} = 0.96 \). Agrees exactly. Why this identity is useful. It finds one ratio from another without finding the angle, which avoids rounding error and gives exact answers when the input is exact. It is also the foundation of the trigonometric identities studied in Algebra 2 and Precalculus. Proved by dividing the Pythagorean theorem by \( c^2 \); \( \cos\theta = 0.96 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is the definition of a circle?
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   All points in a plane at a fixed distance from a fixed center

2. A central angle measures \( 45^\circ \). What is its intercepted arc?
   Show the full solution

   \( 45^\circ \)

3. A radius is 7. Find the diameter.
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   14

4. What is a chord through the center called?
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   A diameter

5. How many times does a tangent meet the circle?
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   Once

6. An arc measures \( 140^\circ \). Find the measure of the rest of the circle.
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   The whole circle measures \( 360^\circ \). \( 360 - 140 = 220^\circ \). The \( 140^\circ \) arc is minor and the \( 220^\circ \) arc is major, since one is under and the other over \( 180^\circ \). \( 220^\circ \)

7. Two radii form a \( 100^\circ \) angle. Find the base angles of the triangle they make with the chord.
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   Both segments are radii, so they are congruent and the triangle is isosceles. By the isosceles triangle theorem the base angles are congruent, and they sum with \( 100^\circ \) to \( 180^\circ \). Each base angle: \( \dfrac{180 - 100}{2} = 40^\circ \). Check: \( 40 + 40 + 100 = 180 \). Correct. \( 40^\circ \) each

8. Three points divide a circle into arcs in the ratio \( 2 : 3 : 4 \). Find all three arc measures.
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   Let the arcs be \( 2x \), \( 3x \) and \( 4x \). They make up the whole circle: \( 2x + 3x + 4x = 360 \), so \( 9x = 360 \) and \( x = 40 \). The arcs measure \( 80^\circ \), \( 120^\circ \) and \( 160^\circ \). Check: \( 80 + 120 + 160 = 360 \). Correct, and the ratio \( 80 : 120 : 160 \) reduces to \( 2 : 3 : 4 \). Correct. \( 80^\circ \), \( 120^\circ \), \( 160^\circ \)

9. Explain why every triangle formed by two radii and a chord is isosceles, and why this matters.
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   By the definition of a circle, every point on it is the same distance from the center. So any two radii have the same length, making them congruent segments. A triangle with two congruent sides is isosceles by definition, so the triangle formed by two radii and the chord joining their endpoints is always isosceles, whatever the central angle. Why it matters. The isosceles triangle theorem then gives two congruent base angles for free, and that pair of congruent angles is usually the missing ingredient in a circle proof. The chord theorems of lesson 10.2 all come from it, the inscribed angle theorem of lesson 10.4 is proved by building an isosceles triangle from a radius, and the tangent results of lesson 10.3 use the same idea. The practical advice. In any circle proof, draw radii to the labeled points. It costs nothing and usually produces the congruent angles or sides the proof needs. All radii of a circle are congruent by definition, so any two of them form an isosceles triangle, supplying congruent base angles

10. Prove that all circles are similar, and explain what follows from it.
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    The proof. Let the first circle have center \( P \) and radius \( r \), and the second have center \( Q \) and radius \( s \). Apply the translation carrying \( P \) to \( Q \). Translation is a rigid motion, so the image is a circle of radius \( r \) centered at \( Q \). Now apply the dilation centered at \( Q \) with scale factor \( k = \dfrac{s}{r} \). A dilation multiplies every distance from the center by \( k \), so a point at distance \( r \) from \( Q \) maps to a point at distance \( r \cdot \dfrac{s}{r} = s \). Every point of the image circle therefore lands at distance \( s \) from \( Q \), and every point at distance \( s \) is the image of some point at distance \( r \). The image is exactly the second circle. A translation followed by a dilation is a similarity transformation, so the circles are similar by the definition of lesson 8.2. Note what was not required. No condition on the two circles was used. Unlike triangles, which need AA or SSS similarity to be checked, circles are automatically similar. The same is true of squares and of regular polygons with the same number of sides, and for the same reason: their shape is determined with no free parameters beyond size. What follows. Corresponding measurements in similar figures are proportional, so any ratio of two lengths within a circle is the same for all circles. That is why \( \dfrac{\text{circumference}}{\text{diameter}} \) has one universal value, named \( \pi \), rather than a value that depends on the circle. It is also why the arc length formula of lesson 10.6 works: the ratio of an arc to the whole circumference depends only on the arc's angular measure, not on the radius. A translation followed by a dilation carries any circle onto any other, so all circles are similar; this is why \( \pi \) is a constant