Appendix A — Answer Key, Chapter 5: Proving Triangles Congruent

SOL G.TR.2 (a, d, e) · Covers textbook Chapter 5 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 126 across the chapter. Proofs show one acceptable order of steps, not the only one.

Conventions used in every answer below: a congruence statement's letter order is part of the claim, so $\triangle ABC \cong \triangle DEF$ asserts $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. CPCTC may be used only after a congruence has been proved. A shared side is congruent to itself by the Reflexive Property. Every proof line carries a claim and a reason.

The five criteria, and the two that are not:

Lesson 5.1 — Congruent Triangles and Corresponding Parts

Guided practice

1. $E$.
2. $\overline{DF}$.
3. $\overline{AB} \cong \overline{DE}$, $\overline{BC} \cong \overline{EF}$, $\overline{AC} \cong \overline{DF}$, $\angle A \cong \angle D$, $\angle B \cong \angle E$, $\angle C \cong \angle F$.
4. The two triangles must already have been proved congruent. CPCTC reads parts off a congruence; it cannot establish one.
5. Because the order names the correspondence. $\triangle ABC \cong \triangle EFD$ claims $A \leftrightarrow E$, $B \leftrightarrow F$, $C \leftrightarrow D$ — a different set of six statements, and usually a false one.
6. Once two triangles are known to be congruent, every pair of matching sides and matching angles is congruent.

Independent practice

7. a) $\overline{PQ}$ b) $\angle R$ c) $\overline{MO}$
8. $DE = 7$, $EF = 9$, $DF = 5$.
9. $m\angle V = 40°$, $m\angle W = 75°$, $m\angle X = 180° - 40° - 75° = 65°$.
10. The angle congruent to $\angle E$ is $\angle B$; the side congruent to $\overline{EF}$ is $\overline{BC}$.
11. $\triangle JKL \cong \triangle RST$.
12. It got $K \leftrightarrow T$ and $L \leftrightarrow R$ wrong; the true matches are $K \leftrightarrow S$ and $L \leftrightarrow T$. ($J \leftrightarrow S$ is also wrong, so strictly all three are misstated, but the two named are the pairs the student swapped.)
13. $\overline{AC}$ corresponds to itself. It is the shared side, congruent to itself by the Reflexive Property, and it counts as one of the three congruences a criterion needs.
14. The corresponding edge of the second tabletop is also $31$ inches, by CPCTC — the tops are congruent because they were cut from one template.
15. $m\angle E = 58°$, by CPCTC.
16. $\overline{AB}$ corresponds to $\overline{DE}$, not $\overline{EF}$. The correct statement is $\overline{AB} \cong \overline{DE}$.
17. Congruence means one triangle can be carried exactly onto the other by rigid motions, which preserve both length and angle measure. A triangle has three sides and three angles, so all six pairs match — not just the sides.
18. CPCTC's hypothesis is that the triangles are congruent. Using it before that is established assumes the conclusion, which makes the argument circular and proves nothing.

Exit ticket 5.1

19. $\angle F$.
20. $YZ = 14$, by CPCTC.
21. Because the order states which vertex matches which, and therefore which sides and angles are being claimed congruent. Change the order and you change the claim.
22. Only after the two triangles have been proved congruent by one of the five criteria.

Lesson 5.2 — SSS and SAS

Guided practice

23. Three pairs of congruent sides — SSS.
24. Two pairs of congruent sides and the pair of angles between them — SAS.
25. $\angle A$.
26. Because $\angle B$'s vertex, $B$, is an endpoint of $\overline{AB}$ but not of $\overline{AC}$. An included angle's vertex must be an endpoint of both marked sides.
27. SSA. It is not a congruence criterion.
28. If two sides of one triangle and the angle included between them are congruent to two sides and the included angle of another, the triangles are congruent.

Independent practice

29. a) SSS b) SAS ($\angle B$ is included between $\overline{AB}$ and $\overline{BC}$) c) none — $\angle A$ is not included between $\overline{AB}$ and $\overline{BC}$, so this is SSA
30. $\angle Y$.
31. $\overline{XZ}$ and $\overline{YZ}$.
32. $\overline{AC} \cong \overline{AC}$, by the Reflexive Property.
33. For SAS: $\angle K \cong \angle N$, the angle included between the two marked sides. For SSS: $\overline{JL} \cong \overline{MP}$.
34. Given $\overline{AB} \cong \overline{CB}$ and $\overline{AD} \cong \overline{CD}$, prove $\triangle ABD \cong \triangle CBD$.

35. Given $\overline{RS} \cong \overline{RU}$ and $\angle SRT \cong \angle URT$, prove $\triangle RST \cong \triangle RUT$.

36. SSS. The two triangles have the two pairs of opposite frame sides congruent, and they share the diagonal, which is congruent to itself.
37. SSS — three pairs of congruent sides ($6$, $6$, $4$ in each).
38. $\angle C$ is not included between $\overline{AB}$ and $\overline{AC}$; its vertex is an endpoint of $\overline{AC}$ but not of $\overline{AB}$. The arrangement is SSA, and no criterion applies.
39. Chapter 4 showed that three lengths either close into a triangle or do not, and when they close there is only one shape — the Triangle Inequality decides existence, and the lengths then fix every angle. So three matching sides force three matching angles.
40. For SSS: one more pair of congruent sides (the shared side plus the given pair makes two, so one more completes three). For SAS: one more pair of congruent angles, specifically the angle included between the two known congruent sides.

Exit ticket 5.2

41. SSS.
42. $\angle B$.
43. SAS — $\angle Q$ is included between $\overline{PQ}$ and $\overline{QR}$, and $\angle Y$ between $\overline{XY}$ and $\overline{YZ}$.
44. The shared side is congruent to itself, by the Reflexive Property of Congruence.

Lesson 5.3 — ASA and AAS

Guided practice

45. Two pairs of congruent angles and the pair of sides between them — ASA.
46. The same two pairs of angles, but the marked side is not between them — AAS.
47. $\overline{AB}$.
48. Because two angles determine the third by the angle sum, so an AAS arrangement can always be rewritten as an ASA arrangement.
49. The shared diagonal gives a congruence of that segment with itself, by the Reflexive Property.
50. The crossing gives a congruent pair of vertical angles, because vertical angles are congruent.

Independent practice

51. a) ASA — $\overline{AC}$ is included between $\angle A$ and $\angle C$ b) AAS — $\overline{AB}$ is not included between $\angle A$ and $\angle C$ c) none — this is AAA, which proves similarity, not congruence
52. $\overline{PR}$.
53. $\angle Q$ and $\angle R$.
54. For ASA: $\overline{XY} \cong \overline{AB}$, the side between the two marked angles. For AAS: any other pair of corresponding sides, for example $\overline{YZ} \cong \overline{BC}$.
55. Given $\overline{AC}$ and $\overline{BD}$ intersect at $E$, $\overline{AE} \cong \overline{CE}$, $\angle BAE \cong \angle DCE$, prove $\triangle ABE \cong \triangle CDE$.

56. Given $\angle B \cong \angle D$ and $\angle BAC \cong \angle DAC$, prove $\triangle ABC \cong \triangle ADC$.

57. AAS. The free given was the shared side $\overline{AC}$, congruent to itself by the Reflexive Property. It is AAS rather than ASA because $\overline{AC}$ is not between $\angle B$ and $\angle BAC$ — it is a side of the angle $\angle BAC$ and opposite $\angle B$.
58. The three angles of each triangle sum to $180°$, so $m\angle C = 180° - m\angle A - m\angle B$ and $m\angle F = 180° - m\angle D - m\angle E$. The two right sides are equal because the corresponding angles are, so $\angle C \cong \angle F$.
59. ASA, using the vertical angles at the crossing as the free given. (The shared segment is the included side between the measured $47°$ angle and the vertical angle.)
60. ASA — the two base angles with the ridge beam included between them.
61. $\overline{AC}$ is not included between $\angle A$ and $\angle B$; the included side would be $\overline{AB}$. The correct criterion is AAS.
62. For angles, the angle sum recovers the missing information: two angles determine the third, so a non-included side is still enough. For sides, nothing plays that role — knowing two sides tells you nothing about the third, and the SSA figure shows two different triangles fitting the same data.

Exit ticket 5.3

63. ASA.
64. AAS. $\overline{PQ}$ has endpoints $P$ and $Q$, but the marked angles are at $P$ and $R$, so it is not included between them.
65. A pair of congruent vertical angles, because vertical angles are congruent.
66. Because the three angles sum to $180°$, so the third is determined once the other two are known.

Lesson 5.4 — HL, and What Does Not Work

Guided practice

67. A right angle, a hypotenuse, and one leg — with the hypotenuses congruent to each other and the marked legs congruent to each other.
68. Because the right angle is part of HL's hypothesis. Without it the given parts form an SSA arrangement, which determines nothing.
69. $\triangle ABC_1$ and $\triangle ABC_2$. They share the angle at $A$ ($30°$), the side $AB = 7$, and the swung side of length $4.5$.
70. Because the swung length $4.5$ is greater than the perpendicular distance from $B$ to the ray but less than $AB$, so the circle of radius $4.5$ about $B$ crosses the ray in two places.
71. All three pairs of angles are congruent. The side lengths are not — one triangle is a scaled copy of the other.
72. Similarity. It is Chapter 7's subject.

Independent practice

73. a) Congruent — HL. b) None — $\angle C$ is not included between $\overline{AB}$ and $\overline{BC}$, so this is SSA. c) None — AAA gives similarity, not congruence. d) Congruent — ASA ($\overline{AB}$ is included between $\angle A$ and $\angle B$).
74. SAS, using the two legs and the included right angle. HL is not needed, and this is simpler.
75. A swung side of $8$ would exceed $AB = 7$, so the arc would cross the ray only once (on the far side of $A$), and the SSA data would determine a single triangle. The ambiguity appears only when the swung side is longer than the perpendicular distance and shorter than $AB$.
76. The arc would be tangent to the ray, meeting it at exactly one point — the foot of the perpendicular. The resulting triangle would be a right triangle, and the data would determine it uniquely. That tangent case is precisely the situation HL covers.
77. Given $\angle B$ and $\angle E$ are right angles, $\overline{AC} \cong \overline{DF}$, $\overline{AB} \cong \overline{DE}$, prove $\triangle ABC \cong \triangle DEF$.

78. HL — the sloped edges are the congruent hypotenuses and the posts are the congruent legs. The posts must be vertical, that is perpendicular to the ground, so that each triangle really contains a right angle.
79. Not congruent. They are similar — same shape, different size.
80. SSA is not a congruence criterion. The SSA figure gives two non-congruent triangles with identical two-sides-and-an-angle data, so the correct conclusion is that nothing follows: the triangles may or may not be congruent.
81. HL requires right triangles, and the right angles were never established. Without them the givens are an SSA arrangement, and no conclusion follows.
82. Congruence is about size as well as shape, and only a side carries size information. Angles alone fix the shape — AAA — and leave the size free, so a criterion made only of angles can never force congruence. Every valid criterion therefore contains at least one S.

Exit ticket 5.4

83. If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and a leg of another right triangle, the triangles are congruent. It applies only to right triangles.
84. SSA and AAA.
85. Because three congruent angles fix only the shape. A scaled copy has the same three angles and different side lengths, so the triangles need not be congruent — they are similar.
86. Because in a right triangle the hypotenuse and one leg determine the other leg exactly, by the Pythagorean Theorem: $b = \sqrt{c^2 - a^2}$ has one non-negative solution. The two possible triangles of the general SSA case collapse to one.

Lesson 5.5 — Direct and Indirect Proofs

Guided practice

87. Given: $\overline{AB} \cong \overline{AD}$ and $\overline{CB} \cong \overline{CD}$. Prove: $\angle B \cong \angle D$.
88. Step 3, $\overline{AC} \cong \overline{AC}$, justified by the Reflexive Property of Congruence.
89. Step 4, and the criterion is SSS.
90. Step 5. It cannot come earlier because CPCTC requires the congruence proved in step 4; using it before that would assume the conclusion.
91. State the assumption, which must be the negation of what is to be proved.
92. Name the contradiction reached, and reject the assumption — concluding the original statement.

Independent practice

93. Given $\overline{AB} \cong \overline{CB}$ and $\overline{AD} \cong \overline{CD}$, prove $\angle A \cong \angle C$.

94. Given $\overline{PQ} \parallel \overline{SR}$ and $\overline{PQ} \cong \overline{SR}$, prove $\triangle PQR \cong \triangle RSP$.

95. a) Reflexive Property of Congruence. b) Vertical angles are congruent. c) CPCTC.
96. Sample paragraph proof for item 93: It is given that $\overline{AB} \cong \overline{CB}$ and $\overline{AD} \cong \overline{CD}$. The two triangles share side $\overline{BD}$, which is congruent to itself by the Reflexive Property. With all three pairs of corresponding sides congruent, $\triangle ABD \cong \triangle CBD$ by SSS. Therefore $\angle A \cong \angle C$ by CPCTC.
97. a) Assume $\triangle ABC \not\cong \triangle DEF$. b) Assume $m\angle A \le 60°$. c) Assume the two lines are not parallel.
98. Assume a triangle has two obtuse angles. Each measures more than $90°$, so the two together measure more than $180°$. But the three angles of a triangle sum to exactly $180°$, and the third angle has positive measure — a contradiction. The assumption is false, so no triangle has two obtuse angles.
99. Assume $\triangle ABC \not\cong \triangle DEF$. It is given that $\overline{AB} \cong \overline{DE}$, $\angle B \cong \angle E$, and $\overline{BC} \cong \overline{EF}$, with $\angle B$ included between the two sides. By SAS the triangles are congruent — contradicting the assumption. The assumption is false, so $\triangle ABC \cong \triangle DEF$.
100. Given two support triangles with $\overline{AB} \cong \overline{DE}$, $\angle A \cong \angle D$, $\overline{AC} \cong \overline{DF}$:

101. CPCTC is used in step 2, but the congruence it depends on is not proved until step 5. The argument is circular: it assumes what it sets out to show.
102. An indirect proof works by ruling the assumption out, and the only thing that rules it out is an impossibility. Without a contradiction the assumption remains perfectly possible, so nothing has been eliminated and nothing established.

Exit ticket 5.5

103. The step naming the congruence criterion and concluding that the triangles are congruent.
104. The Reflexive Property of Congruence.
105. Assume $\triangle PQR \not\cong \triangle STU$.
106. It must state the assumption — the negation of the conclusion — in its first line, and name the contradiction it reaches in its last, then reject the assumption.

Lesson 5.6 — Constructing a Congruent Triangle, and Measured Attributes

Guided practice

107. One side of the original is copied: a ray is drawn and a segment congruent to $\overline{AB}$ is laid off on it with the compass.
108. They locate the third vertex, $C'$. Each arc is the set of points at the correct distance from one endpoint, so a crossing point is at the correct distance from both — which is exactly what the third vertex must be.
109. SSS — all three side lengths were copied.
110. Because G.TR.2d assesses the construction, not the finished drawing. The arcs are the evidence that the third vertex was located with a compass rather than by eye or by tracing.
111. SSS — the tick marks show all three pairs of sides congruent.
112. CPCTC.

Independent practice

113. (1) Draw a ray and use the compass to lay off a segment congruent to one side of the original, giving $\overline{A'B'}$. (2) From $A'$, swing an arc with radius equal to a second side. (3) From $B'$, swing an arc with radius equal to the third side. (4) Label the crossing point $C'$ and draw $\overline{A'C'}$ and $\overline{B'C'}$, leaving the arcs on the page.
114. Copy one side to get $\overline{A'B'}$. At $A'$, copy the angle of the original at $A$, producing a ray. On that ray, lay off a segment congruent to the original's second side, giving $C'$. Draw $\overline{B'C'}$. The triangle is congruent by SAS.
115. $DE = 8$, $EF = 13$, $FD = 10$.
116. $m\angle M = 38°$, $m\angle N = 96°$, $m\angle P = 180° - 38° - 96° = 46°$.
117. $DF = 17$ m and $m\angle E = 72°$, both by CPCTC.
118. $\overline{XY}$ corresponds to $\overline{RS}$, not $\overline{RT}$. So $XY = 6$; the side equal to $12$ is $\overline{XZ}$.
119. Sides $4$ m, $6$ m, and $7$ m, and an apex angle of $80°$ — identical to the first, by CPCTC.
120. The longest distance is $55$ m, the side of her triangle corresponding to the longest side across the ravine. The reason is CPCTC, applied after the congruence is established.
121. $\angle C$ corresponds to $\angle F$, not $\angle E$. The correct statement is $m\angle F = 41°$.
122. The arithmetic afterwards is trivial — you simply copy a number across. All the risk sits in matching the right parts, and writing the correspondence out forces that match to be made explicitly instead of guessed from position on the page.

Exit ticket 5.6

123. Copy one side of the original: draw a ray and lay off a congruent segment with the compass.
124. Because they are the evidence that the construction was performed, and G.TR.2d assesses the construction itself. Without them a correct construction cannot be told from a traced drawing.
125. $EF = 21$, by CPCTC.
126. Establish the congruence with one of the five criteria; write the correspondence in order; apply CPCTC to the part asked for; answer in context with units.

Chapter 5 Review

Review 1 (G.TR.2a).

• The crossing gives $\angle AEB \cong \angle CED$, because vertical angles are congruent.
• Proof:

• $\overline{AB} \cong \overline{CD}$ by CPCTC. It could not have come earlier because CPCTC requires the congruence established in step 4; placing it before that would assume the conclusion.
• With $\overline{AB} \cong \overline{CD}$ in place of $\overline{BE} \cong \overline{DE}$, the known parts of $\triangle ABE$ are the sides $\overline{AE}$ and $\overline{AB}$ and the angle $\angle AEB$. That angle's vertex is $E$, which is an endpoint of $\overline{AE}$ but not of $\overline{AB}$, so it is not included between them: the arrangement is SSA, and nothing follows. A replacement that does work: $\angle A \cong \angle C$ gives $\overline{AE}$ included between $\angle A$ and $\angle AEB$, so ASA applies. ($\angle B \cong \angle D$ also works, by AAS.)

Review 2 (G.TR.2a).

(i) SSS. (ii) None — SSA. (iii) AAS. (iv) None — AAA gives similarity. (v) HL.

Why (ii) fails but (v) succeeds: in a general triangle, two sides and a non-included angle leave the third vertex free to land in either of two places, so two non-congruent triangles fit the same data. In a right triangle the Pythagorean Theorem pins the third side down — $b = \sqrt{c^2 - a^2}$ has exactly one non-negative value — so the ambiguity disappears and the data determines the triangle.

Review 3 (G.TR.2 d, e).

• The SSS construction: draw a ray and lay off a segment congruent to one side of the template; from one endpoint swing an arc equal to a second side; from the other endpoint swing an arc equal to the third side; the arcs cross at the third vertex; draw the two remaining sides. The arcs must be left on the page, because they are the evidence the construction was performed.
• The $12$-inch side of the bracket corresponds to the $12$-inch side of the template, by CPCTC. The other two sides are $9$ in and $15$ in.
• HL and SSS both apply — HL because the right angle is present and the hypotenuse ($15$ in, opposite the right angle) plus one leg would suffice, and SSS because all three sides are known. (SAS also applies, using the two legs and the included right angle.)
• Measure the template's angle opposite its $9$-inch side once. Every bracket is congruent to the template, so by CPCTC the corresponding angle in each bracket equals that measurement. Measuring every bracket is unnecessary because congruence transfers all six parts at once — one measurement of the template settles that angle for the whole production run.

Workbook-only items

Page 2, fill in the blanks. Congruent triangles have six pairs of congruent parts: three pairs of sides and three pairs of angles. The order of the letters in a congruence statement is part of the claim.

Page 3, CPCTC frame. CPCTC stands for Corresponding Parts of Congruent Triangles are Congruent. It may be used only after the triangles have been proved congruent.

Page 5, criteria frame. SSS — three pairs of sides. SAS — two pairs of sides and the included angle. ASA — two pairs of angles and the included side. AAS — two pairs of angles and a non-included side. HL — hypotenuse and one leg, right triangles only.

Page 6, included-angle blank. An angle is included between two sides when its vertex is an endpoint of both. Mark the two sides first, then look at where the angle is.

Page 9, free-givens frame. A shared side is congruent to itself by the Reflexive Property. Two segments that cross give a congruent pair of vertical angles.

Page 12, the two that fail. SSA — two different triangles fit the same data. AAA — fixes shape but not size; this is similarity, covered in Chapter 7. Every valid criterion contains at least one S.

Page 14, proof-shape frame. Write the givens; add the free congruence; name the criterion; use CPCTC for the part asked for.

Page 15, indirect-proof frame. The first line must state the assumption, the negation of the conclusion. The last line must name the contradiction and reject the assumption.

Page 17, construction steps. (1) Copy one side. (2) Swing an arc from each endpoint with the lengths of the other two sides. (3) The arcs cross at the third vertex. (4) Draw the remaining two sides and keep the arcs.

Page 19, four-step method. Establish the congruence; write the correspondence in order; apply CPCTC; answer in context with units.

Pages 7, 16, and 18, blank frames. Any assigned proof, marked figure, or construction. Expected conventions: matching tick marks and arcs for matching parts; every proof line carrying both a claim and a reason; the criterion named before CPCTC is used; construction arcs left visible.