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CHAPTER 2 TRIANGLES
Section 2.1 Classifying Triangles
2.1 PRACTICE EXERCISES
+
1. In ABC , m∠C = 90° and
AC ≠ BC ≠ AB ≠ AC.
(a) Since m∠C = 90°, ∠C is a right angle so
ABC is a right triangle.
(b) Since all three sides are of unequal length,
the triangle is scalene.
+
(c) Sides AC and BC include ∠C since the
vertex of ∠C , C , is on both AC and BC.
(d) ∠B is opposite side AC.
(e) Side AB is included between ∠A
and ∠B.
(f) Side BC is opposite ∠A.
(g) The hypotenuse of a right triangle is the
side opposite the right angle. Since AB is
opposite the right angle ∠C , AB is the
hypotenuse.
(h) The sides of a right triangle that are not
the hypotenuse are called its legs. Thus, the
legs of ABC are AC and BC. The legs
include the right angle, ∠C.
+
2. Let x be the length of each side of an
equilateral triangle with perimeter 75 ft.
Then
x + x + x = 75
3 x = 75
x = 25
Thus each side is 25 ft long.
2.1 SECTION EXERCISES
1. D, E, and F
11. AB, BC , and AC
2. DF , FE , and DE
12. A, B, and C
3. Obtuse triangle
13. Equilateral triangle
4. Isosceles triangle
14. Equiangular triangle
5. DE
15. BC
6. DF
16. AC
7. ∠F
17. ∠A
8. ∠D
18. ∠C
9. DE
19. It has no hypotenuse since it is not a right
triangle.
10. It has no hypotenuse since it is not a right
triangle.
20. Yes
21
22
CHAPTER 2 TRIANGLES
21. 20 + 30 + 40 = 90; 90 cm
37. False ( ∠CED is not adjacent to ∠GEF )
22. 54 ft
38. True
23. 33 in (12.5 + 12.5 + 8 = 33)
39. True
24. 30 cm (Solve 14 + 22 + x = 66.)
40. False (the angles are supplementary)
25. 23 ft (Solve x + x + x = 69.)
41. Yes, a triangle can be scalene and obtuse.
Scalene means all sides have different lengths
and obtuse means the triangle has one obtuse
angle.
26. 17 in (Solve x + x + 13 = 47.)
27. base: 15 cm, sides: 45 cm
1
(Solve x + x + x = 105.)
3
28. base: 6 ft, sides: 12 ft
1
(Solve x + x + x = 30)
2
29.
+AED and +AEB
+ ABD and +DCB
31. +ECD and + ACD
32. + ABD and + ABE
42. No, scalene means all sides of the triangle
have different lengths but isosceles means
two sides have the same measure.
43. Yes, a right triangle can be isosceles. A right
triangle contains one right angle while an
isosceles triangle has two congruent sides.
30.
33. True
44. Yes, an isosceles triangle has two sides
congruent and an acute triangle has three
acute angles.
34. True
35. False (neither side of the angle is a side of the
triangle)
36. True
45. No, an equilateral triangle has three angles
that measure 60°.
Section 2.2 Congruent Triangles
23
Section 2.2 Congruent Triangles
2.2 PRACTICE EXERCISES
1. We can show the desired congruence in two ways.
First, since AC = 3 cm and US = 3 cm,
Second, since AB = 5 cm and TS = 5 cm,
AC ≅ SU . Also, we are given that CB ≅ UT ,
and that ∠C and ∠U are right angles
making ∠C ≅ ∠U . Then ABC = STU by
SAS
AB ≅ ST . Also, since AC = 3 cm and US = 3 cm,
+
+
AC ≅ SU , and since we are also given that
CB ≅ UT ,
+ABC ≅7STU
by SSS
2.2 SECTION EXERCISES
1. Congruent by SSS
5. Congruent by ASA
2. Congruent by ASA
6. Since vertical ∠ 's are ≅ , the triangles are ≅
by ASA
3. Since right angles are congruent, the triangles
are congruent by SAS
4. Congruent by SSS
8.
7. 1. Given
3. BD ≅ BD
2. ∠3 ≅ ∠4
4. ASA
C is midpoint of AE
∠E ≅ ∠ A
∠ACB ≅ ∠ECD
Given
Given
Vertical ∠ 's ≅
2
↓
→
AC ≅ EC
+ABC ≅+EDC
Def. of midpoint
9. 1. Given
10. 1. Given
ASA
2. ⊥ lines form rt. ∠ 's
2. AC ≅ BD
0
3. ∠ABD ≅ ∠CDB
3. BC ≅ BC
4. DB ≅ BD
5. Given
6. SAS
4. SSS
Note to students about proofs: Proofs are not unique. Your proof may differ slightly from the solutions
manual. This does not mean your proof is necessarily incorrect. Consult with your instructor.
11. Proof: STATEMENTS
REASONS
1. AD bisects BE
1. Given
2. BC ≅ EC
2. Def. of bisector
3. BE bisects AD
3. Given
4. AC ≅ DC
5. ∠ACB ≅ ∠DCE
6. ABC ≅ DEC
4. Def. of bisector
5. Vertical angles are ≅
6. SAS
+
+
24
CHAPTER 2 TRIANGLES
12. Proof:
STATEMENTS
REASONS
1. ∠B and ∠E are right angles
1. Given
2. m∠B = m∠E
2. Rt. ∠ 's are ≅
3. AD bisects BE
3. Given
4. BC ≅ EC
4. Def. of bisector
5. ∠BCA ≅ ∠ECD
5. Vert. ∠ 's are ≅
6. ∠ABC ≅ ∠DEC
6. ASA
13. Proof: STATEMENTS
REASONS
1. AD ≅ BD so AD = BD
1. Given; Def. ≅ seg.
2. AE ≅ BC so AE = BC
3. AD − AE = BD − BC
4. ED = AD − AE and CD = BD − BC
2. Given; Def. ≅ seg.
3. Add.-Subt. Post.
4. Seg-Add. Post.
5. ED = CD so ED ≅ CD
6. ∠D ≅ ∠D
7. ACD ≅ BED
5. Substitution; Def. ≅ seg
6. Reflexive Law
7. SAS
+
14. Proof:
+
STATEMENTS
1. DB ⊥ AC
1. Given
2. ∠DBA and ∠DBC are right angles
2.
3. ∠DBA ≅ ∠DBC
3. Rt. ∠ 's are ≅
4. DB bisects AC
4. Given
5. AB ≅ CB
5. Def. of bisector
6. DB ≅ DB
6. Reflexive law
7.
7. SAS
+ABD ≅+CBD
15. (a) Not correct because BA 
≅ ED
(b) Correct
(c) Not correct because ED 
≅ AB
(d) Not correct, BAC ≅ FED by SAS
+
16.
REASONS
+
+ ACB ≅+QPR by SAS or ASA
+ DEF ≅ + MLN by SAS or SSS
+ DEF ≅ + GJH by SSS
+ MLN ≅ + GJH by SSS
⊥ lines form rt. ∠ 's
17. (b) Same
(c) Match
(d) Match; yes (e) Yes
(f) No, no AA Theorem because the
corresponding sides of two triangles may
not match even if corresponding angles are
congruent.
18. Group Activity
Section 2.3 Proofs Involving Congruence
19. Since AC = PR, we must solve
x + 1 = 3x − 5.
1 = 2 x − 5 Subtract x from both sides
6 = 2x
Add 5 to both sides
3= x
Divide both sides by 2
20. Since m∠B = m∠Q, we must solve
100 + y = 5 y + 20
100 = 4 y + 20 Subtract y from
both sides
80 = 4 y
Subtract 20 from
both sides
20 = y
21. Since AC = x + 1 and x = 3 (by Exercise 19),
substituting 3 for x we obtain
AC = 3 + 1 = 4
25
22. Since PR = 3x − 5 and x = 3 (by Exercise
19), substituting 3 for x we obtain
PR = 3(3) − 5 = 9 − 5 = 4.
23. Since m∠B = (100 + y )° and y = 20 (by
Exercise 20), substituting 20 for y we obtain
m∠B = (100 + 20)° = 120°.
24. Since m∠Q = (5 y + 20)° and y = 20 (by
Exercise 20), substituting 20 for y we obtain
m∠Q = (5(20) + 20)° = (100 + 20)° = 120°.
25. PR
26. BA
27. QR
28. ∠B
29. ∠P
30. ∠C
Section 2.3 Proofs Involving Congruence
6. 1. Given 2. ∠1 ≅ ∠2 3. Adj. ∠ 's whose
noncommon sides are in a line are supp. 4.
∠2 and ∠4 are supplementary 5. ∠3 ≅ ∠4
6. Def. of vert. ∠ 's 7. ∠GCB ≅ ∠FCD
8. GBC ≅ FDC 9. CPCTC
1. 9 cm
2. 12 cm
+
3. 46°
4. 102°
5. 1. Given 2. ∠1 ≅ ∠2 3. Adj. ∠ 's whose
noncommon sides are in a line are supp. ∠ 's
4. ∠2 and ∠ECB are supplementary
5. Supp. of ≅ ∠ 's are ≅ 6. Reflexive Law
7. SAS 8. ∠G ≅ ∠E
+
7. 1. Given 2. Given 3. AB ≅ CD; AB = CD
4. Seg.-Add. Post. 5. BD = CD + BC
7. Substitution law; Def. ≅ seg.
8. AFC ≅ DEB 9. CPCTC
U
8. 1. Given 2. Given 3. ∠ABE ≅ ∠EBC
4. BC ≅ BE 5. ASA 6. ∠A ≅ ∠D
Note to students about proofs: Proofs are not unique. Your proof may differ slightly from the solutions
manual. This does not mean your proof is necessarily incorrect. Consult with your instructor.
9. Proof:
STATEMENTS
1. AC ≅ CE
REASONS
1. Given
2. DC ≅ CB
3. ∠DCE and ∠BCA are vertical angles
4. ∠DCE ≅ ∠BCA
5. DCE ≅ BCA
2. Given
3. Def. of vert. ∠ 's
4. Vert. ∠ 's ≅
5. SAS
6. ∠A ≅ ∠E
6. CPCTC
+
+
26
CHAPTER 2 TRIANGLES
10. Proof:
1.
2.
3.
4.
STATEMENTS
¶
l
AC bisects ∠BAD
∠1 ≅ ∠2
¶
l
CA bisects ∠BCD
∠3 ≅ ∠4
5. AC ≅ CA
6. 7 ABC ≅7 ADC
7. ∠B ≅ ∠D
11. Proof:
REASONS
1. Given
2. Def. of ∠ bisector
3. Given
4. Def. of ∠ bisector
5. Reflexive law
6. ASA
7. CPCTC
STATEMENTS
REASONS
1. BC ≅ CD
2. ∠1 ≅ ∠2
1. Given
2. Given
3. AC ≅ AC
4. ABC ≅ ADC
3. Reflexive Law
4. SAS
5. AB ≅ AD
5. CPCTC
U
12. Proof:
U
¶l
STATEMENTS
REASONS
1. DB bisects ∠ADC
2. ∠1 ≅ ∠2
1. Given
2. Def. of ∠ bisector
3. AD ≅ CD
3. Given
4. DB ≅ DB
4. Reflexive law
5. SAS
5.
+ABD ≅+CBD
6. ∠ABD ≅ ∠CBD
6. CPCTC
7. ∠ABD and ∠CBD are adjacent angles
7. Def. of adj. ∠ 's
8. DB ⊥ AC
8. Def. of ⊥ lines
13. Proof:
STATEMENTS
REASONS
1. ∠1 ≅ ∠2
2. ∠3 ≅ ∠4
1. Given
2. Given
3. DB ≅ BD
4. ABD ≅ CDB
5. ∠A ≅ ∠C
3. Reflexive Law
4. ASA
5. CPCTC
+
+
Section 2.3 Proofs Involving Congruence
14. Proof:
27
STATEMENTS
REASONS
1. BC ⊥ AB
1.
Given
2. ∠ABC is a right angle
2.
⊥ lines form rt. ∠ 's
3. AD ⊥ DC
3.
Given
4. ∠ADC is a right angle
4.
⊥ lines form rt. ∠ 's
5. ∠ABC ≅ ∠ADC ; m∠ABC = m∠ADC
5.
All rt. ∠'s are ≅; and Def. ≅ ∠ ' s
6. ∠1 ≅ ∠2, m∠1 = m∠2
6.
Given; Def. ≅ ∠ ' s
7. m∠ABC − m∠1 = m∠ADC − m∠2
7.
Add.-Subtr. Post.
8. m∠3 = m∠ABC − m∠1
8.
∠ Add. Post.
9. ∠4 = m∠ADC − m∠2
9.
∠ Add. Post.
10. m∠3 = m∠4 so ∠3 ≅ ∠4
10. Substitution and Def. ≅ ∠ ' s
11. BD ≅ DB
11. Reflexive law
12.
12. ASA
+ABD ≅+CDB
13. AB ≅ CD
15. Proof:
STATEMENTS
13. CPCTC
REASONS
1. GB ⊥ AF and FD ⊥ GE
1. Given
2. ∠1 and ∠2 are right angles
2. ⊥ lines form rt. ∠ 's
3. ∠1 ≅ ∠2
3. Rt. ∠ 's are ≅
4. GD ≅ FB and GB ≅ FD
4. Given
5.
5. SAS
+BGF ≅+DFG
6. ∠BGF ≅ ∠DFG; m∠BGF = m∠DFG
6. CPCTC; Def. ≅ ∠ 's.
7. ∠3 ≅ ∠4; m∠3 = m∠4
7. CPCTC; Def. ≅ ∠ 's.
8. m∠DFG − m∠3 = m∠BGF − m∠4
8. Add.-Subt. Post.
9. m∠5 = m∠DFG − m∠3
9. ∠ Add. Post.
10. m∠6 = m∠BGF − m∠4
10. ∠ Add. Post.
11. m∠5 = m∠6 so ∠5 ≅ ∠6
11. Substitution; Def ≅ ∠ 's.
12.
12. ASA
+BGC ≅+DFC
13. BC ≅ DC
13. CPCTC
28
CHAPTER 2 TRIANGLES
16. Proof:
STATEMENTS
REASONS
1. ∠A ≅ ∠C and ∠1 ≅ ∠2; m∠1 = m∠2
1.
Given; Def. ≅ ∠ 's
2. B is the midpoint of AC
2.
Given
3. AB ≅ BC
4. ABG ≅ CBD
3.
Def. of midpoint
4.
ASA
5. BG ≅ BD
6. ∠GBD ≅ ∠GBD; m∠GBD = m∠GBD
5.
CPCTC
6.
Reflexive law; Def. ≅ ∠ ' s
7. m∠1 + m∠GBD = m∠2 + m∠GBD
7.
Add.-Subtr. Post.
8. m∠CBF = m∠2 + m∠GBD
8.
Angle Add. Post.
9. ∠ABE = m∠1 + m∠GBD
9.
Angle Add. Post.
10. m∠CBF = m∠ABE so ∠CBF ≅ ∠ABE
10. Substitution and Def. ≅ ∠ ' s
11.
11. ASA
+
+
+CBF =+ABE
12. BF ≅ BE
13. GF = BF − BG and DE = BE − BD
13. Seg.-Add. Post.
14. GF =DE so GF ≅ DE
14. Add.-Subt. Post.; Def. ≅ seg.
17. Since AB and DE are corresponding parts of
congruent triangles ABC and EDC (by
SAS), AB = 105 yd (the same as DE).
+
+
18. For stability. The triangle is a rigid figure that
cannot be distorted like a four-sided figure.
The triangle is formed by three fixed
distances and it is the only such triangle
possible by SSS.
19. Because the triangle is a rigid figure that
cannot be distorted like a four-sided figure.
This is a result of SSS since there is only one
triangle possible with three given sides. The
bridge cannot change shape without breaking.
20. Since the triangle formed by her eyes, Q, and
P is congruent to the triangle formed by her
eyes, S, and P (by ASA, one angle is a right
angle), PQ ≅ PS since they are
corresponding sides in congruent triangles.
Thus, PQ, the width of the canyon, is also 45
ft.
12. CPCTC
21. AD ≅ CD and ∠ADB ≅ ∠CDB are given.
DB ≅ DB by reflexive law, thus
ADB ≅ CDB by SAS and AB ≅ CB by
CPCTC.
8x − 3 = 6 x +1
2x = 4
x=2
+
+
22. Given AC ≅ DF thus 4 x + 2 = 3x + 3
x =1
23. ∠ZWY ≅ ∠XYW and ∠ZYW ≅ ∠XWY are
given. WY ≅ YW by reflexive law, thus
ZYW ≅ XWY by ASA and WZ ≅ YX by
CPCTC.
+
+
3x − 2 = 2 x + 1
x=3
Section 2.4 Isosceles Triangles, Medians, Altitudes and Concurrent Lines
24. AB ≅ DB and EB ≅ CB are given.
∠ABE ≅ ∠DBC because vertical ∠ ' s are ≅
thus ABE ≅ DBC by SAS and AE ≅ DC
by CPCTC
2x − 6 = x
x=6
+
+
29
25. ∠BAC ≅ ∠DCA and ∠CAD ≅ ∠ACB is
given information, AC ≅ CA by reflexive
law, thus ABC ≅ CDA by ASA.
AB ≅ CD by CPCTC.
+
+
Section 2.4 Isosceles Triangles, Medians, Altitudes and Concurrent Lines
2.4 PRACTICE EXERCISES
1. The reason for Statement 1 is: Given
Statement 2 is: AE ≅ DE
+
This follows since in AED, ∠A ≅∠D
making the sides opposite ∠A and ∠D, AE
and ED, congruent.
Statement 3 is: ∠1 ≅ ∠2
This follows since the only other given
information not already used is ∠1 ≅ ∠2 .
The reason for Statement 4 is: ASA
The reason for this is because ∠1 ≅ ∠2,
∠A ≅ ∠D, and the included sides AE and
DE are also congruent.
The reason for Statement 5 is: CPCTC
Note that BE and CE correspond in
congruent triangles AEB and DEC .
+
Statement 6 is:
+
+BCE is isosceles
This follows since an isosceles triangle has
two congruent sides and BE ≅ CE.
k
l
2. 1. ED because it is the perpendicular bisector
of BC.
+
+
2. AD because it is a median of ABC.
k
l
3. AF because it is an altitude of ABC.
k
l
4. BG because it is an angle bisector of
∠ABC.
2.4 SECTION EXERCISES
2. AB ≅ AE
3. ∠ 's opp. ≅ sides are ≅
1. AB ≅ CB
8. 1. Given
2. ∠M ≅ ∠N
5. CPCTC
3. WY ≅ XY
6. ∠3 ≅ ∠4
9. 1. Given
2. AC ≅ AD so AC = AD
3. Given
4. Def. of midpoint and Def. ≅ seg.
4. ∠Q ≅ ∠R
5. RS ≅ TR
6. XZ ≅ YZ
7. 1. Given 2. AC bisects ∠BAD
3. Def. of ∠ bisector 4. Reflexive law
5. BAE ≅ DAE 6. CPCTC
7. BDE is isosceles
+
+
4. ASA
+
5. E is the midpoint of AD
6. Def. of midpoint and Def. ≅ seg.
7. Seg. Add. Post. 8. Substitution law
10. Mult.-Div. Post. 13. ∠1 ≅ ∠2
10. 1. Given 2. ∠1 ≅ ∠2 3. BD ≅ CE
4. ABD ≅ AEC 5. CPCTC
+
+
30
CHAPTER 2 TRIANGLES
11. Proof:
STATEMENTS
REASONS
1. ∠1 ≅ ∠2
1. Given
2. BE ≅ CE
2. Sides opp ≅ ∠ 's are ≅
3. ∠3 ≅ ∠4
3. Given
4. ∠AEB and ∠DEC are vertical angles
4. Def. of vert. ∠ 's
5. ∠AEB ≅ ∠DEC
5. Vert. ∠ 's are ≅
6.
6. ASA
+ABE ≅+DCE
7. ∠A ≅ ∠D
12. Proof:
7. CPCTC
STATEMENTS
REASONS
1. AB ≅ AE
1. Given
2. ∠E ≅ ∠B
2. Angles opp. ≅ sides are ≅
3. BC ≅ DE
3. Given
4.
4. SAS
+ABC ≅+AED
5. AC ≅ AD
5. CPCTC
6. ∠1 ≅ ∠2
6. Angles opp. ≅ sides are ≅
13. An altitude is a line segment from a vertex of
a triangle, perpendicular to the side opposite
the vertex (or an extension of that side). What
distinguishes this from the other two is an
altitude of a triangle may be outside the
triangle and an altitude is perpendicular to a
side of the triangle.
An angle bisector is a ray that separates an
angle into two congruent adjacent angles.
What distinguishes this from the other two is
all angle bisectors of a triangle are inside the
triangle and cuts the angle into two congruent
angles.
A median of a triangle is a line segment
joining a vertex with the midpoint of the
opposite side of the triangle. What
distinguishes this from the other two is all
medians of a triangle are inside the triangle
and the midpoint of a side of the triangle is
needed.
14. True
15. True
16. False
17. True
18. False
19. False
20. Since F is the centroid, AF is
2
( AF + FI )
3
2
AF = ( AF + 3)
AF
3
2
AF = AF + 2 or 3 AF
3
3 AF
1
AF = 2
AF
3
AF = 6 units
2
( AI ). Then
3
AF =
2
( AF + FI )
3
= 2( AF + 3)
= 2 AF + 6
= 6 units
=
21. The orthocenter of a triangle is the
intersection of the altitudes of the triangle. From
the figure we see this point is D.
Section 2.5 Proving Right Triangles Congruent
22. In an equilateral triangle, the incenter,
circumcenter, orthocenter and centroid all
coincide.
2
23. AF = ( AI )
3
2
AF = (10.5)
3
AF = 7
FI = AI − AF
FI = 10.5 − 7
FI = 3.5 inches
28. The circumcenter is the point of concurrency
needed. The circumcenter is equidistant from
the angles of a triangle. This was observed in
the group activity and technological insights
for the perpendicular bisecors of a triangle.
29. No, the centroid is on a median of a triangle.
The median joins a vertex with the midpoint
of the opposite side. A median is always
inside a triangle.
2
( AF + FI )
3
= 2( AF + 6.8)
= 2 AF + 13.6
= 13.6
= AF + FI
= 13.6 + 6.8
= 20.4 m
AF =
3 AF
3 AF
AF
AI
AI
AI
+
26. AT is a median of ABC since it contains a
centroid. Point I is the midpoint of BC
therefore CI = IB.
2x + 3 = x + 5
x=2
27. By Theorem 2.12, the bisectors of the angles
of a triangle meet at a point equidistant from
the sides of a triangle. The incenter is the
point of intersection of the angle bisectors,
thus point E is the desired point.
2
24.
AF = ( AI )
3
2
10 = ( AI )
3
15 mm. = AI
25.
31
30. Yes, the orthocenter bisectors can be outside
an obtuse triangle. An orthocenter is the point
of concurrency of the altitudes of a triangle.
In an obtuse triangle, some altitudes are
outside of the triangle.
Section 2.5 Proving Right Triangles Congruent
2.5 PRACTICE EXERCISES
1. 3. Since both triangles contain right angles, they are right triangles by the definition of a right triangle.
4. Given
5. Since B is the midpoint of AC , definition of a midpoint states it separates it into two congruent parts.
6. Statements 4 and 5 show two pairs of legs in the right triangles are congruent, thus the triangles are
congruent by LL.
2.5 SECTION EXERCISES
1. LA
2. LL
+ ABD, + ABC
6. + ABT , + ADT
4.
8. Given
3. LA
10. ⊥ lines form rt. ∠ 's
5.
11. Def. rt
+ABT
+
12. LL
7. Given
13. Given
14. Given
9. Given
15. Reflexive Law
16. SSS
32
CHAPTER 2 TRIANGLES
Note to students about proofs: Proofs are not unique. Your proof may differ slightly from the solutions
manual. This does not mean your proof is necessarily incorrect. Consult with your instructor.
17.
STATEMENTS
REASONS
1. XY ≅ YZ
1. Given
2. WY ≅ WY
2. Reflexive Law
3. WY ⊥ XZ
4. ∠XYW and ∠ZYW are rt. ∠ 's
3. Given
4. ⊥ lines form rt ∠ 's
5.
6.
5. Def. rt
6. LL
+XYW and +ZYW
+XYW ≅ +ZYW
18.
are rt.
STATEMENTS
+
REASONS
1. C is midpoint AE and BD
1. Given
2. AC ≅ ED
3. ∠ACB ≅ ∠ECD
4. ABC ≅ EDC
2. Def. of midpt
3. Vertical ∠ 's ≅
4. SAS
+
19.
+
STATEMENTS
REASONS
1. AB ≅ DE
2. ∠ACB ≅ ∠ECD
1. Given
2. Vertical ∠ 's ≅
3. AB ⊥ BD, DE ⊥ BD
3. Given
4. ∠ABC and ∠EDC are rt ∠ 's
4. ⊥ lines form rt ∠ 's
5.
6.
5. Def. rt
6. LA
+ABC and +EDC
+ABC ≅ +EDC
are rt
7. AC ≅ EC
+
7. CPCTC
20. STATEMENTS
REASONS
1. BD ≅ CE
2. ∠A ≅ ∠A
1. Given
2. Reflexive law
3. BD ⊥ AC ; CE ⊥ AB
3. Given
4. ∠AEC and ∠ADB are rt ∠ 's
4. ⊥ form rt ∠ 's .
5.
6.
5. Def rt.
6. LA
+AEC and + ADB are rt
+AEC ≅+ADB
7. AE ≅ AD
7. CPCTC
Section 2.6 Constructions Involving Triangles
33
Section 2.6 Constructions Involving Triangles
2.6 PRACTICE EXERCISES
1. We use Construction 2.2. First construct an
angle equal to ∠A by using Construction 1.2.
Place the compass point at the vertex of this
angle and mark off lengths equal to BC on
each of its sides. Then draw the segment
formed by these points to make the desired
DEF .
U
2.6 SECTION EXERCISES
1. Use Construction 2.1.
2. Use Construction 2.1.
3. Use Construction 2.2.
4. Use Construction 2.2.
5. Use Construction 2.3.
6. Use Construction 2.3.
7. Any two obtuse angles can be tried since their
noncommon sides will not intersect.
8. Many examples will work.
18. Use Construction 2.5 three times. The point of
concurrency is the orthocenter.
19. Use Construction 2.6 three times. The point of
concurrency is the centroid.
20. Use Construction 2.6 three times. The point of
concurrency is the centroid.
21. Use Construction 1.6 three times. The point of
concurrency is the incenter.
22. Use Construction 1.6 three times. The point of
concurrency is the incenter.
11. The sides do not intersect to form a triangle.
23. No, the orthocenter is the point of
concurrency of the altitudes of a triangle.
Some altitudes in an obtuse triangle will be
outside the triangle, thus, so will the
orthocenter.
12. First construct a right angle using
Construction 1.4. Then use Construction 2.2.
24. Yes, the centroid is the point of concurrency
of the medians of a triangle.
13. First construct a right angle using
Construction 1.4, then use Construction 2.2.
25. Yes, the incenter is the point of concurrency
of the angle bisectors.
14. Use Construction 1.4 then bisect the right
angle using Construction 1.6.
26. Circumcenter is the point of concurrency of
the perpendicular bisectors.
15. There are two triangles with these parts.
27. They intersect at the circumcenter.
16. A triangle cannot be formed since the 1-inch
side will not reach the other side.
28. No, not if the sum of the two smaller lengths
is less than the larger length.
17. Use Construction 2.5 three times. The point of
concurrency is the orthocenter.
29. Use Construction 2.4.
9. Use Construction 2.2.
10. Use a construction similar to Construction
1.1.
30. The 3 medians intersect at the centroid.
The 3 altitudes intersect at the orthocenter.
The 3 angle bisectors intersect at the incenter.
34
CHAPTER 2 TRIANGLES
Chapter 2 Review Exercises
1. A, B, and C
11. True
2. AB, BC, and CA
12. True
3. Scalene triangle
13. False ( ∠FEG is not adjacent to ∠BED )
4. Obtuse triangle
14. True
5. AC
15. Yes
6. ∠B
16. SSS
17. ASA
7. BC
18. 1. Given 2. AC ≅ AE
4. ACF = AEB
+
8. ∠C
9. 44 cm (44 = 12 + 16 + 16)
+
3. ∠A ≅ ∠A
19. 1. Given 2. AC bisects ∠BAD
3. ∠BAC ≅ ∠EAC 4. Reflexive Law
5. ABC = AEC
10. base: 13 ft, sides: 26 ft
1
(Solve x + x + x = 65.)
2
+
+
Note to students about proofs: Proofs are not unique. Your proof may differ slightly from the solutions
manual. This does not mean your proof is necessarily incorrect. Consult with your instructor.
20. Proof:
STATEMENTS
REASONS
1. AD ≅ CB
1. Given
2. AB ≅ CD
2. Given
3. DB ≅ BD
4. ADB ≅ CBD
3. Reflexive Law
4. SSS
+
21. Proof:
+
STATEMENTS
REASONS
1. AC ⊥ BD
2. ∠3 ≅ ∠4
3. ∠1 ≅ ∠2
1. Given
2. ⊥ lines form ≅ adj. ∠ 's
3. Given
4. EC ≅ EC
5. DEC ≅ BEC
4. Reflexive Law
5. ASA
6. DC ≅ BC
6. CPCTC
7. AC ≅ AC
8. ABC = ADC
7. Reflexive Law
8. SAS
9. AB ≅ AD
9. CPCTC
+
+
+
+
Chapter 2 Review Exercises
22. Since AC ≅ EF by ≅
35
, we must solve
x + 2 = 4 x − 4.
2 = 3 x − 4 Subtract x from both sides
6 = 3x
Add 4 to both sides
2=x
23. Since ∠E ≅ ∠A by ≅
, we must solve
y + 10 = 2 y − 15
10 = y − 15 Subtract y from both sides
25 = y
Add 15 to both sides
24. Since AC = x + 2 and x = 2 (from Exercise
22), substituting 2 for x we obtain
25. Since ∠A = (2 y − 15)° and y = 25 (from
Exercise 23), substituting 25 for y we obtain
∠A = (2(25) − 15)° = (50 − 15)° = 35°.
26. DF corresponds to BC.
27. ∠B corresponds to ∠D.
¶
l
28. 1. Given 2. AC bisects ∠DAB 3. Def. of
∠ bisector; Def. ≅ ∠ 's 4. Given
5. ∠EBA ≅ ∠DBE so m∠EBA = m∠DBE
6. ∠ Add. Post. 7. Substitution law
9. Sym. and trans. laws
10. Mult.-Div. Post. and Def ≅ ∠ 's
11. Reflexive law 12. ASA 13. AC ≅ BE
AC = 2 + 2 = 4.
29. Proof:
STATEMENTS
REASONS
1. E is the midpoint of AC
1. Given
2. AE ≅ CE
2. Def. of midpoint
3. E is the midpoint of BD
3. Given
4. BE ≅ DE
4. Def. of midpoint
5. ∠AEB and ∠CED are vertical angles
5. Def. of vert. ∠ 's
6. ∠AEB ≅ ∠CED
6. Vert. ∠ 's are ≅
7.
7. SAS
+AEB ≅+CED
8. AB ≅ CD
30. Proof:
8. CPCTC
STATEMENTS
REASONS
1. AB ≅ CD
1. Given
2. BC ≅ DE
2. Given
3. ∠CAE ≅ ∠CEA
3. Given
4. AC ≅ CE
4. Sides opp. ≅ ∠ 's are ≅
5.
5. SSS
+ABC ≅+CDE
6. ∠B ≅ ∠D
6. CPCTC
36
CHAPTER 2 TRIANGLES
31. Proof:
STATEMENTS
1.
34. False
35. True
36. True
37. False
38. False
1. Given
2. AB ≅ AD
2. Def. of isosceles
3.
3. Given
+BDE is isosceles with base BD
4. BE ≅ DE
4. Def. of isosceles
5. AE ≅ AE
5. Reflexive law
6.
6. SSS
+ABE ≅+ADE
+
+
7. ∠3 ≅ ∠4
7. CPCTC
8. ∠3 and ∠1 are supplementary and
∠4 and ∠2 are supplementary
8. Adj. ∠ 's whose noncommon sides are
in line are supp.
9. ∠1 ≅ ∠2
9. Supp. of ≅ ∠ ' s are ≅
32. Proof:
33. True
+ABD is isosceles with base BD
REASONS
STATEMENTS
REASONS
1. AB ≅ CB; AD ≅ CD
1. Given
2. BD ≅ BD
2. Reflexive law
3.
3. SSS
+ABD ≅+CBD
4. ∠ABD ≅ ∠CBD
4. CPCTC
5. BD is bisector of ∠ABC
5. Def. ∠ bisector
39.
CE = 2 (CE + EF )
3
17 = 2 (17 + EF )
3
3(17) = 2(17 + EF )
51 = 34 + 2( EF )
17 = 2( EF )
8.5 m = EF
Since D is the midpoint of AC, AD = CD.
Since AD = 12 m, then CD =12 m.
AC = 12 + 12 = 24 m
AC = 24 m
Chapter 2 Practice Test
40. Proof:
37
STATEMENTS
REASONS
1. ∠C ≅ ∠E
1. Given
2. AD ⊥ BC ; AD ⊥ DE
2. Given
3. ∠ABC and ∠BDE are rt ∠ 's
3. ⊥ lines form rt ∠ 's
4.
4. Def. rt
+ABC , and +BDE are rt
5. B is midpoint AD
5. Given
6. AB ≅ DB
7. ABC ≅ BDE
6. Def. of midpoint
7. LA
+
41. Proof:
+
STATEMENTS
1.
+WXY
REASONS
is isosceles triangle with WX ≅ YX
1. Given
2. ∠W ≅ ∠Y
2. Sides opp. ≅ ∠ 's are ≅ .
3. XZ ≅ XZ
3. Reflexive law
4. WY ⊥ XZ
4. Given
5. ∠XZW , ∠XZY are rt ∠ 's.
5. ⊥ lines form rt ∠ 's.
+WXZ and +YXZ are rt
7. +WXZ ≅+YXZ
6.
.
6. Def. rt.
7. LA
42. Use Construction 2.1.
44. Use Construction 2.6.
43. First construct a right angle using
Construction 1.4. Then use the right angle, the
segment, and the acute angle with
Construction 2.3.
45. Use Construction 2.2.
46. Use Construction 2.5.
Chapter 2 Practice Test
1. Acute triangle
8. Yes
2. Isosceles triangle
9. ∠ACD
3. ∠A
4. ∠B ≅ ∠C
10. 13 cm (AB = AC so AC = 5 cm; then
13 = 5 + 5 + 3)
5. ∠B
11. 33 in (all three sides measure 11 inches).
6. BC
12. 28° (Since ∠C ≅ ∠D, solve x + 20 = 48
giving x = 28.)
7. Yes
38
CHAPTER 2 TRIANGLES
13. Proof:
STATEMENTS
REASONS
1. ∠1 ≅ ∠2
1. Given
2. D is the midpoint of CE
2. Given
3. CD ≅ DE
3. Def. of midpoint
4. AC ≅ AE
4. Given
5. ∠E ≅ ∠C
5. Angles opp. ≅ sides are ≅
6.
6. ASA
+BCD ≅+FED
7. BD ≅ FD
14. Proof:
7. CPCTC
STATEMENTS
REASONS
1. ∠3 ≅ ∠4
1. Given
2. AC ≅ AD
2. Sides opp. ≅ ∠ 's are ≅
3. ∠1 and ∠3 are supplementary and
∠2 and ∠4 are supplementary
3. Adj. ∠ 's whose concommon sides
are in line are supp.
4. ∠1 ≅ ∠2
4. Supp. of ≅ ∠ 's are ≅
5. BC ≅ ED
5. Given
6.
6. SAS
+ABC ≅+AED
7. ∠5 ≅ ∠6
15. Proof:
STATEMENTS
7. CPCTC
REASONS
1. AB bisects CD
1. Given
2. DE ≅ CE
2. Def. bisector
3. ∠DEB ≅ ∠CEA
3. Vertical ∠ 's ≅
4. ∠C and ∠D are rt. ∠ 's
4. Given
+ACE and +BDE are rt
6. + ACE ≅+BDE
5.
5. Def. rt
6. LA
+
Chapter 2 Practice Test
16. Use Construction 2.2 followed by
Construction 2.5 and Construction 2.6.
39
20. (a) centroid
(b)
17. It is both an altitude and a perpendicular
bisector because it is perpendicular to the base
and it bisects
as indicated by the symbol
the base because the marks show the
segments on both sides of the bold line are
congruent. It is an altitude because the
segment has an endpoint at a vertex of the
triangle.
18. The bold segment is a perpendicular bisector
but not an altitude. It is perpendicular and
bisects one side of the triangle. This segment
is not an altitude because an endpoint is not at
a vertex of the triangle.
19. Neither
2
( PW + WX )
3
2
29 = (29 + WX )
3
2
3(29) = 3< (29 + WX )
3
87 = 2(29 + WX )
87 = 58 + 2(WX )
29 2(WX )
=
2
2
14.5 m = WX
PW =
(c) MX = 9 m since M is the midpoint of
MN (because PX is a median.)