Geometry Honors Notes – Chapter 4: Congruent Triangles – Solutions to Proof Practice Problems
4.3 – Prove Triangles Congruent by SSS
PRACTICE #1
4.4 – Prove Triangles Congruent by SAS and HL
PRACTICE #1
D
E
Given: AD ≅ CD;
Given: ∠1 ≅ ∠2;
B is the midpoint of AC.
Prove: ΔABD ≅ ΔCBD
M is the midpt. of BC.
BE ≅ CE
Prove: ΔEMB ≅ ΔEMC
A
Statements
1.
2.
3.
4.
5.
C
B
Reasons
AD ≅ CD
B is the midpoint of AC.
AB ≅ CB
BD ≅ BD
ΔABD ≅ ΔCBD
1.
2.
3.
4.
5.
Statements
O
5. M is the midpt. of BC.
6. MB ≅ MC
7. ΔEMB ≅ ΔEMC
N
Given: PR ≅ NT , NO ≅ SR;
O is 13 of the way from N to P.
S is
1
3
of the way from R to T.
Prove: ΔNRT ≅ ΔRNP
Statements
1.
2.
PRACTICE #2
S
T
Given: BC ⊥ AC ,
BD ⊥ AD,
Reasons
1
3
C
B
A
AC ≅ AD
Prove: ΔACB ≅ ΔADB
1. Given
PR ≅ NT
NO ≅ SR,
2. Given
1
O is 3 of the way from N to P.
S is
Statements
D
Reasons
of the way from R to T.
3.
NP ≅ RT
4.
5.
NR ≅ NR
ΔNRT ≅ ΔRNP
R
C D
1. Given
2. Given
3. If 2 ∠s form a st. ∠
(assumed from
diagram), then they are
supp.
4. Congruent
Supplements Th.
5. Given
6. Def. of midpt.
7. SAS (1, 4, 6)
4. ∠3 ≅ ∠ 4
P
4 2
M
Reasons
1. BE ≅ CE
2. ∠1 ≅ ∠2
3. ∠1 is supp. to ∠3
∠2 is supp. to ∠4
Given
Given
Def. of midpoint.
Reflexive Property
SSS (1, 3, 4)
PRACTICE #2
1 3
A B
1. BC ⊥ AC , BD ⊥ AD
2. ∠ACB is a rt. ∠.
∠BDA is a rt. ∠.
3. ΔACB and ΔADB
are right triangles.
4. AC ≅ AD
5. AB ≅ AB
6. ΔACB ≅ ΔADB
3. Multiplication
Prop.
4. Reflexive Prop.
5. SSS (1, 3, 4)
Page 1
1. Given
2. Def. of perpendicular lines.
3. Def. of right triangles.
4. Given
5. Reflexive Property
6. HL (3, 4, 5)
Geometry Honors Notes – Chapter 4: Congruent Triangles – Solutions to Proof Practice Problems
PRACTICE #3
4.5 – Prove Triangles Congruent by ASA & AAS
B
PRACTICE #1
Given: S and T trisect RV ,
∠R ≅ ∠V ,
∠BST ≅ ∠BTS
Prove: ΔBRS ≅ ΔBVT
R
Given: ∠3 ≅ ∠6,
KR ≅ PR,
∠KRO ≅ ∠PRM
Prove: ΔKRM ≅ ΔPRO
R
3 4
K
Statements
1. ∠3 ≅ ∠6
2. ∠3 is supp. to ∠4.
∠6 is supp. to ∠5.
3.
4.
5.
6.
7.
8.
∠ 4 ≅ ∠5
KR ≅ PR
∠KRO ≅ ∠PRM
∠MRO ≅ ∠MRO
∠KRM ≅ ∠PRO
ΔKRM ≅ ΔPRO
S
V
T
5 6
O
M
Statements
P
1.
2.
3.
4.
5.
∠R ≅ ∠V
S and T trisect RV
Reasons
Given
Given
RS ≅ ST ≅ TV
Def. of trisection pts.
Given
∠BST ≅ ∠BTS
∠BSR is supp. to ∠BST .
If 2 ∠s form a st. ∠
∠BTV is supp. to ∠BTS .
(assumed from diag.),
then they are supp.
6. Congruent Supp. Th.
6. ∠BSR ≅ ∠BTV
7. ΔBRS ≅ ΔBVT
7. ASA (1, 3, 6)
Reasons
1. Given
2. If 2 ∠s form a st. ∠
(assumed from
diagram), then they are
supp.
3. Congruent Supp. Th.
4. Given
5. Given
6. Reflexive Property
7. Subtraction Property
8. ASA (3, 4, 7)
1.
2.
3.
4.
5.
4.6.1 – Use Congruent Triangles (CPCTC)
PRACTICE #1
G
PRACTICE #2
A
D
Given: ∠1 ≅ ∠6,
C
BC ≅ EC
Prove: ΔABC ≅ ΔDEC
3 4
1
2
5
B
E
Statements
1. ∠1 ≅ ∠6
2. ∠1 is supp. to ∠2.
∠6 is supp. to ∠5.
3.
4.
5.
6.
∠ 2 ≅ ∠5
BC ≅ EC
∠3 ≅ ∠ 4
ΔABC ≅ ΔDEC
6
G
F
Reasons
m∠GOH = 67°
Answer: m∠GHO = 25°
GH = 13.5
1. Given
2. If 2 ∠s form a st. ∠
(assumed from
diagram), then they are
supp.
3. Congruent Supp. Th.
4. Given
5. Vertical angles are ≅ .
6. ASA (3, 4, 5)
4 x − 105 = x + 24
x = 43
2 y − 7 = 3 y − 23
y = 16
GH = 27 / 2 = 13.5
H
J
G
H
Given: H is the midpt. of GJ .
G
H
M is the midpt. of OK .
H
GO ≅ JK , GJ ≅ OK ,
M
O
∠G ≅ ∠K , OK = 27,
m∠GOH = x + 24, m∠GHO = 2 y − 7,
m∠JMK = 3 y − 23, m∠MJK = 4 x − 105.
Find: m∠GOH , m∠GHO, and GH.
Page 2
K
Geometry Honors Notes – Chapter 4: Congruent Triangles – Solutions to Proof Practice Problems
PRACTICE #2
PRACTICE #4
E
Given: ∠AEC ≅ ∠DEB,
Given:
BE ≅ CE ,
∠ABE ≅ ∠DCE
Prove: AB ≅ CD
A
Statements
1.
2.
3.
4.
5.
6.
7.
∠ABE ≅ ∠DCE
BE ≅ CE
∠AEC ≅ ∠DEB
∠BEC ≅ ∠BEC
∠AEB ≅ ∠DEC
ΔAEB ≅ ΔDEC
AB ≅ CD
C
B
FB ≅ DE ,
∠CFB ≅ ∠AED
Prove: ∠FAB ≅ ∠ECD
D
B
Reasons
H
PRACTICE #5
B
1
K
2
4
J
D
PA ≅ PD,
AB ⊥ FC ,
Reasons
P
A
F
E
DE ⊥ FC.
Prove: ∠C ≅ ∠F
1. Given
2. Given
3.
4.
5.
6.
7.
8.
9.
C
Given: BC ≅ FE ,
3
PF ≅ PC ,
Statements
1. ∠2 ≅ ∠4
2. ∠1 is comp. to ∠2.
∠3 is comp. to ∠4.
3. ∠1 ≅ ∠3
4. KG ≅ GJ
5. ∠FGJ ≅ ∠HGK
6. ∠KGJ ≅ ∠KGJ
7. ∠FGK ≅ ∠HGJ
8. ΔFGK ≅ ΔHGJ
9. FG ≅ HG
G
F
E
A
1. FB ≅ DE
1. Given
2. Given
2. ∠CFB ≅ ∠AED
3. ∠CFB is supp. to ∠BFA. 3. If 2 ∠s form a st. ∠
∠AED is supp. to ∠DEC.
(assumed from diag.),
then they are supp.
4. ∠BFA ≅ ∠DEC
4. Congruent Supp. Th.
5. Given
5. AE ≅ FC
6. Reflexive Property
6. EF ≅ EF
7. AF ≅ EC
7. Addition Property
8. ΔFAB ≅ ΔECD
8. SAS (1, 4, 7)
9. ∠FAB ≅ ∠ECD
9. CPCTC
Given
Given
Given
Reflexive Property
Subtraction Property
ASA (1, 2, 5)
CPCTC
PRACTICE #3
Given: KG ≅ GJ ,
∠2 ≅ ∠4,
∠1 is comp. to ∠ 2.
∠3 is comp. to ∠ 4.
∠FGJ ≅ ∠HGK
Prove: FG ≅ HG
C
F
Statements
Reasons
1.
2.
3.
4.
5.
6.
7.
D
AE ≅ FC ,
Statements
Congruent Comp. Th.
Given
Given
Reflexive Property
Subtraction Property
ASA (3, 4, 7)
CPCTC
1. AB ⊥ FC , DE ⊥ FC
2. ∠BAC is a rt. ∠.
∠EDF is a rt. ∠.
3. ΔBAC and ΔEDF
are right triangles.
4. BC ≅ FE
5. PA ≅ PD
6. PF ≅ PC
7. DF ≅ AC
8. ΔBAC ≅ ΔEDF
9. ∠C ≅ ∠F
Page 3
Reasons
1. Given
2. Def. of ⊥ lines.
3. Def. of right triangles.
4.
5.
6.
7.
8.
9.
Given
Given
Given
Addition Property
HL (3, 4, 7)
CPCTC
Geometry Honors Notes – Chapter 4: Congruent Triangles – Solutions to Proof Practice Problems
PRACTICE #3
4.6.2 – Steps Beyond CPCTC
Given: AZ ≅ ZB;
E
PRACTICE #1
Given: G is the midpt. of FH
EF ≅ EH
Prove: ∠1 ≅ ∠2
W
Z is the midpt. of XY .
∠AZX ≅ ∠BZY ,
1
XW ≅ YW
Prove: AW ≅ BW
2
H
1.
2.
3.
4.
5.
6.
7.
8.
G is the midpt. of FH
FG ≅ HG
EF ≅ EH
Draw EG
EG ≅ EG
ΔEFG ≅ ΔEHG
∠EFG ≅ ∠EHG
∠2 is supp. to ∠EFG.
∠1 is supp. to ∠EHG.
9. ∠1 ≅ ∠2
Z
F
G
Reasons
1.
2.
3.
4.
5.
6.
7.
8.
1.
2.
3.
4.
5.
6.
7.
8.
Given
Def. of midpt.
Given
Two pts. det. a segment.
Reflexive Property
SSS (2, 3, 5)
CPCTC
If 2 ∠s form a st. ∠
(assumed from
diagram), then they are
supp.
9. Congruent Supp. Th.
Statements
A
Statements
1.
2.
3.
4.
5.
6.
7.
∠AEB ≅ ∠DEC
AE ≅ DE
∠A ≅ ∠D
ΔAEB ≅ ΔDEC
AB ≅ CD
BC ≅ BC
AC ≅ BD
1.
2.
3.
4.
5.
6.
7.
8.
JJJG
Given: DF bisects ∠CDE .
JJJG
EF bisects ∠CED.
G is the midpt. of DE.
DF ≅ EF
Prove: ∠CDE ≅ ∠CED
E
AE ≅ DE ,
∠ A ≅ ∠D
Prove: AC ≅ BD
AZ ≅ ZB
∠AZX ≅ ∠BZY
Z is the midpt. of XY .
XZ ≅ ZY
ΔAZX ≅ ΔBZY
AX ≅ BY
XW ≅ YW
AW ≅ BW
Reasons
B
C
Draw FG.
DF ≅ EF
G is the midpt. of DE.
DG ≅ EG
FG ≅ FG
ΔFDG ≅ ΔFEG
∠FDG ≅ ∠FEG
JJJG
DF bisects ∠CDE .
JJJG
EF bisects ∠CED.
9. ∠CDE ≅ ∠CED
1.
2.
3.
4.
5.
6.
7.
8.
D
Reasons
1.
2.
3.
4.
5.
6.
7.
Given
Given
Given
Def. of midpt.
SAS (1, 2, 4)
CPCTC
Given
Subtraction Property
PRACTICE #4
PRACTICE #2
Given: ∠AEB ≅ ∠DEC,
Y
X
Statements
Statements
B
A
Given
Given
Given
ASA (1, 2, 3)
CPCTC
Reflexive Property
Addition Property
Page 4
C
F
E
D
G
Reasons
1.
2.
3.
4.
5.
6.
7.
8.
Two pts. det. a segment.
Given
Given
Def. of midpt.
Reflexive Property
SSS (2, 4, 5)
CPCTC
Given
9. Multiplication Property
Geometry Honors Notes – Chapter 4: Congruent Triangles – Solutions to Proof Practice Problems
4.6.3 – Overlapping Triangles
4.6.4 – Detour Proofs
PRACTICE #1
PRACTICE #1
Y
Z
Given: YW bisects AX .
∠ A ≅ ∠X ,
∠5 ≅ ∠ 6
Prove: ZW ≅ YW
Statements
1.
2.
3.
4.
5.
6.
7.
8.
∠A ≅ ∠X
YW bisects AX .
AW ≅ XW
∠5 ≅ ∠ 6
∠ZWY ≅ ∠ZWY
∠AWY ≅ ∠XWZ
ΔAWY ≅ ΔXWZ
ZW ≅ YW
5
A
6
HJJG
Given: PQ bisects YZ .
X
W
1.
2.
3.
4.
5.
6.
7.
8.
Statements
HJJG
1. PQ bisects YZ .
2.
3.
4.
5.
6.
7.
8.
A
BD ≅ CD;
B
E is the midpt. of YZ .
Prove: ∠BYZ ≅ ∠CZY
Statements
8. YD ≅ ZD
9. ∠BDY ≅ ∠CDZ
10. BD ≅ CD
11. ΔBYD ≅ ΔCZD
12. ∠BYZ ≅ ∠CZY
Y
D
C
E
Z
Reasons
1.
2.
3.
4.
5.
6.
7.
W
P
Y
Q
To reach the required conclusion, we must prove that
___ ΔWQP ≅ ΔXQP ______________, but the given
information is not sufficient to prove these triangles
congruent. Therefore, we must detour through another
pair of triangles _ ΔZWP ≅ ΔYXP __________.
Given
Given
Def. of segment bisector.
Given
Reflexive Property
Addition Property
ASA (1, 3, 6)
CPCTC
Given: YD ≅ ZD,
Draw DE.
E is the midpt. of YZ .
YE ≅ ZE
DE ≅ DE
YD ≅ ZD
ΔEYD ≅ ΔEZD
∠EYD ≅ ∠EZD
Q is the midpt. of WX .
∠Y ≅ ∠Z , WZ ≅ XY
Prove: ∠WQP ≅ ∠XQP
Reasons
PRACTICE #2
1.
2.
3.
4.
5.
6.
7.
Z
Two pts det. a segment.
Given
Def. of midpt.
Reflexive Property
Given
SSS (3, 4, 5)
CPCTC
8. Given
9. Vertical angles are ≅ .
10. Given
11. SAS (8, 9, 10)
12. Addition Property
Page 5
ZP ≅ PY
∠Z ≅ ∠Y
WZ ≅ XY
ΔZWP ≅ ΔYXP
WP ≅ PX
Q is the midpt. of WX .
WQ ≅ QX
9. PQ ≅ PQ
10. ΔWQP ≅ ΔXQP
11. ∠WQP ≅ ∠XQP
Reasons
1. Given
2.
3.
4.
5.
6.
7.
8.
Def. of segment bis.
Given
Given
SAS (2, 3, 4)
CPCTC
Given
Def. of midpt.
9. Reflexive Property
10. SSS (6, 8, 9)
11. CPCTC
X
Geometry Honors Notes – Chapter 4: Congruent Triangles – Solutions to Proof Practice Problems
Alternative Proof:
PRACTICE #2
P
Given: PR ≅ PU ,
Statements
QR ≅ QU ,
RS ≅ UT
Prove: ∠1 ≅ ∠2
1
S
R
Statements
2
Q
T
Reasons
1. Two pts det. a segment.
2. PQ ≅ PQ
2. Reflexive Property
PR ≅ PU
QR ≅ QU
ΔPQR ≅ ΔPQU
∠R ≅ ∠ U
RS ≅ UT
ΔPRS ≅ ΔPUT
∠1 ≅ ∠2
3.
4.
5.
6.
7.
8.
9.
Given
Given
SSS (2, 3, 4)
CPCTC
Given
SAS (3, 6, 7)
CPCTC
Given: PT ≅ PU ,
PR ≅ PS
JJJG
Prove: PQ bisects ∠RPS .
R
PT ≅ PU
∠SPT ≅ ∠RPU
PR ≅ PS
ΔSPT ≅ ΔRPU
∠T ≅ ∠U
∠RQT ≅ ∠SQU
S
T
U
7. Subtraction Property
8. AAS (6, 5, 7)
9. RQ ≅ SQ
9. CPCTC
10. PQ ≅ PQ
12. ΔRPQ ≅ ΔSPQ
13. ∠RPQ ≅ ∠SPQ
JJJG
14. PQ bisects ∠RPS .
10. Reflexive Property
12. SSS (3, 9, 10)
13. CPCTC
14. Def. of angle bisector.
T
Given: ∠3 ≅ ∠4,
BX ≅ AY ,
BW ≅ AZ
Prove: ΔWTZ is isosceles.
Given
Reflexive Property
Given
SAS (1, 2, 3)
CPCTC
Vertical angles are ≅ .
7. RT ≅ SU
8. ΔRTQ ≅ ΔSUQ
11. CPCTC
12. SAS (3, 5, 11)
13. CPCTC
PRACTICE #1
Reasons
1.
2.
3.
4.
5.
6.
1.
2.
3.
4.
5.
6.
7.
4.7 – Use Isosceles and Equilateral Triangles
Q
Statements
1.
2.
3.
4.
5.
6.
11. RQ ≅ SQ
12. ΔRPQ ≅ ΔSPQ
13. ∠RPQ ≅ ∠SPQ
JJJG
14. PQ bisects ∠RPS .
P
PRACTICE #3
PT ≅ PU
∠SPT ≅ ∠RPU
PR ≅ PS
ΔSPT ≅ ΔRPU
∠PRU ≅ ∠PST
∠T ≅ ∠U
RT ≅ SU
Given
Reflexive Property
Given
SAS (1, 2, 3)
CPCTC
CPCTC
Subtraction Property
(1-3)
8. ∠TRQ is supp. to ∠PRU . 8. If 2 ∠s form a st. ∠
(assumed from diag.),
∠USQ is supp. to ∠PST .
then they are supp.
9. ∠TRQ ≅ ∠USQ
9. Congruent Supp. Th.
10. ΔRTQ ≅ ΔSUQ
10. ASA (6, 7, 9)
U
1. Draw PQ.
3.
4.
5.
6.
7.
8.
9.
1.
2.
3.
4.
5.
6.
7.
Reasons
B
W
Statements
1. ∠3 ≅ ∠4
2. ∠3 is supp. to ∠WBX .
∠4 is supp. to ∠YAZ .
3.
4.
5.
6.
7.
8.
14. Def. of angle bisector.
Page 6
∠WBX ≅ ∠ZAY
BX ≅ AY
BW ≅ AZ
ΔWBX ≅ ΔZAY
∠W ≅ ∠Z
ΔWTZ is isosceles.
3
X
4
Y
A
Z
Reasons
1. Given
2. If 2 ∠s form a st. ∠
(assumed from diag.),
then they are supp.
3. Congruent Supp. Th
4. Given
5. Given
6. SAS (4, 3, 5)
7. CPCTC
8. If at least two ∠s of a
Δ are ≅ , the triangle is
isosceles.
Geometry Honors Notes – Chapter 4: Congruent Triangles – Solutions to Proof Practice Problems
PRACTICE #2
PRACTICE #3
C
Given: CE ≅ CF ,
∠F ≅ ∠3;
∠E is supp. to ∠5.
Prove: ΔCDG is isosceles.
D 3
4 G
5
E
Statements
Given: ΔABC is isosceles with AB ≅ AC.
D is the midpt. of AB.
E is the midpt. of AC.
D
Prove: ΔPBC is isosceles.
F
CE ≅ CF
∠E ≅ ∠F
∠F ≅ ∠3
∠E ≅ ∠3
∠4 is supp. to ∠5.
1.
2.
3.
4.
5.
6.
7.
8.
9.
∠E is supp. to ∠5.
∠E ≅ ∠4
∠3 ≅ ∠4
ΔCDG is isosceles.
6.
7.
8.
9.
P
E
B
Reasons
1.
2.
3.
4.
5.
A
Statements
1. ΔABC is isosceles
with AB ≅ AC.
2. D is the midpt. of AB.
3. E is the midpt. of AC.
4. DB ≅ EC
5. ∠ABC ≅ ∠ACB
6. BC ≅ BC
7. ΔDBC ≅ ΔECB
8. ∠DCB ≅ ∠EBC
9. ΔPBC is isosceles.
Given
Base Angles Theorem
Given
Transitive Property
If 2 ∠s form a st. ∠
(assumed from diag.),
then they are supp.
Given
Congruent Supp. Th.
Transitive Property
If at least two ∠s of a
Δ are ≅ , the triangle is
isosceles.
1. Given
2.
3.
4.
5.
6.
7.
8.
9.
Given
Given
Division Property
Base Angles Theorem
Reflexive Property
SAS (4, 5, 6)
CPCTC
If at least two ∠s of a Δ
are ≅ , the triangle is
isosceles.
Alternative Proof:
Statements
1. ΔABC is isosceles
with AB ≅ AC.
2. ∠A ≅ ∠A
3. D is the midpt. of AB.
4. E is the midpt. of AC.
5. AD ≅ AE
6. ΔABE ≅ ΔACD
7. ∠ABE ≅ ∠ACD
8. ∠ABC ≅ ∠ACB
9. ∠PBC ≅ ∠PCB
10. ΔPBC is isosceles.
Page 7
C
Reasons
Reasons
1. Given
2. Reflexive Property
3. Given
4. Given
5. Division Property
6. SAS (1, 2, 5)
7. CPCTC
8. Base Angles Theorem
9. Subtraction Property
10.If at least two ∠s of a
Δ are ≅ , the triangle is
isosceles.