### Use CPCTC

```p.244ex4
G
E
SAS. Steps 1,3,4
Reflex. Prop of
F
H
Given
Alt. Int. <s Thm.
Given
Statement
Reason
J
M
p.246ex4
K
Statement
L
Reason
Reflex. Prop of ≅
Given
Given
SAS Steps 1,2,3
<JKL≅<MLK
p.244 ex 4
B
C
A
D
Statement
Reason
Given
Given
Alt. Int. <s Thm.
SAS Steps 3,2,4
Reflexive Prop of ≅
p.247: 21
Given: <ZVY≅<WYV, <ZVW≅<WYZ,VW≅YZ
V
W
Prove:
X
Y
Statement
Z
Reason
<ZVY≅<WYV, <ZVW≅<WYZ
m <ZVY + m <ZVW = m <WYV + m <WYZ
m <ZVY = m <WYV,
m <ZVW = m <WYZ
<WVY≅ <ZYV
Given
Def. of ≅
Def. of ≅
Given
Reflex. Prop of ≅
SAS, Steps 6,5,7
m<WVY = m<ZYV
Determine if you can use ASA to prove the triangles congruent. Explain.
No, no included side
p. 246:13
A
Given: B is the midpoint of
Statement
Reason
D
B
C
Given
Given
B is the mdpt of DC
Def. Mdpt.
SAS Steps
2,4,5
Reflex. Prop
of ≅
<ABD and <ABC are
right <s
<ABD≅<ABC
X
Determine if you can use ASA to prove ΔUVX≅ΔWVX. Explain.
Statement
Reason
U
V
p.253ex2
W
given
Def. of Linear
Pair
<WVX ≅ <UVX
<WVX is a
right angle
<UXV ≅ <WXV
Reflex. Prop
given
Given:
What is the measure of y?
l
1000
y
m
p.253ex2
Determine if you can use ASA to prove ΔNKL≅ΔLMN. Explain.
Reflex. Prop
K
L
<KLN≅<MNL
By Alt. Int. <s Thm,
N
M
No other congruence relationships can be
determined, so ASA cannot be applied.
p.255ex4
Determine is you can use the HL Congruence Theorem to prove the triangles congruent.
If not, tell what else you need to know.
Yes
Yes
No, need the hyp ≅
Since <ABC and <DCB are rt <s, ΔABC
and ΔDCB are rt triangles.
Seg. CB ≅ Seg. CB, by the Reflexive Prop.
It is given that segment AC ≅ segment DB.
ΔABC≅ΔDCB by HL.
L
Given: <G≅ <K, <J≅<M, HJ≅LM
Prove: ΔGHJ≅ΔKLM
p.254ex3
H
K
G
ΔGHJ ≅ ΔKLM
M
J
Given
Given
Statement
Reason
ASA Steps
1,3,2
Third <s Thm
<H ≅ <L
<G ≅ <K, <J ≅ <M
p.254ex3
Use AAS to prove the triangles congruent.
Y
Given: <X ≅ <V, <YZW ≅ <YWZ,
Prove: ΔXYZ≅ΔVYW
X
Z
W
V
<X ≅ <V
Statement
Reason
<YZX ≅ <YWV
≅XYZ ≅ ΔVYW
<YZW ≅ <YWZ
AAS
Given
≅ Supps Thm
Given
Given
Def. of Supp <s
Def. of Supp <s
<XZY is supp
to <YZW
<YWX is supp to
<VWY
p. 257: 13
A
F
B
Given:
Prove:
C
Statement
E
D
Reason
Rt. < ≅Thm
AAS
Given
Given
Given
p.257:15
Given: E is a midpoint of Segments AD and BC
Prove: Triangles ABE and DCE are congruent
A
B
E
C
Statement
Reason
D
<A and <D are rt angles
Given
Given
HL
Rt. <s Thm
Def. of mdpt
Def. Rt Δs
E is mdpt of
p.258: 22
A
B
Given:
Prove:
E
D
Statement
C
Reason
AAS
Given
Vert. <s Thm
Alt. Int. <s Thm
Given
p. 258: 23
J
Given:
K
Prove:
M
L
Statement
Reason
Rt.<s Thm
Given
Given
Def. of Perpendicular
AAS
B
Given:
p.259q4
D
Prove:
F
Statement
A
C
E
G
Reason
Def. of Supp <s
Given
<BAC is supp
of <FAB;
<DEC is supp
of <GED
ASA
Given
≅ Supp Thm
Given:
Use CPCTC
Prove:
E
G
Statement
D
F
Reason
Converse of Alt. Int. <s Thm
Given
Reflex. Prop of ≅
Given
Alt. Int. <s Thm
CPCTC
SAS
Prove:
O
Use CPCTC
M
Statement
p.261ex3b
N
Given:
P
Reason
AAS
Alt. Int. <s Thm.
CPCTC
Conv. Alt. Int. <s Thm
Reflex. Prop ≅
Given
C
Given:
Use CPCTC
B
Prove:
A
D
Statement
Reason
Given
SSS
Reflex. Prop of ≅
CPCTC
Def. of < Bisector
p.263: 8
Given:
M is the midpoint of
Prove:
Use CPCTC
M
P
Statement
Q
R
S
Reason
Def. of mdpt
CPCTC
Given
SAS
Vert <s Thm
W
Given:
p.263: 9
X
Use CPCTC
Prove:
Z
Statement
Y
Reason
CPCTC
Reflex. Prop ≅
SSS
Given
p.263: 10
Given:
G is the midpoint of
E
G is the midpoint of
Use CPCTC
Through any 2 points
there is exactly 1 line
Prove:
1
F
2
G
Given
≅ Supp. Thm
H
Reflex. Prop of ≅
Statement
Reason
Draw
SSS
CPCTC
Given
Def. of mdpt
Def. of ≅
FG = HG
p.263: 11
Given:
L
Use CPCTC
Prove:
M is the midpoint of
J
CPCTC
K
M
Given
Statement
Reason
Reflex. Prop of ≅
Def. of < bisector
M is the midpoint of
Def. of mdpt
SAS
Given
p.263:14
Given: ΔQRS is adjacent to ΔQTS.
Prove:
Statement
Reason
Def. of bisect
Given
Def. of < bisect
AAS
CPCTC
Reflex. Prop of ≅
p.263: 15
Given:
with E the midpoint of
Prove:
Use CPCTC
Conv. of Alt. Int. <s Thm
SAS
Statement
Reason
Given
CPCTC
Def. of mdpt
Vert <s Thm
E is the mdpt. of
P
Given: PS = RQ, m<1 = m<4
Prove: m<3 = m<2
p.264:19
Q
3
4
Use CPCTC
1
S
2
R
Def. of Perpendicular
Reflex. Prop of ≅
Def. of rt triangle
m<1 = m<4
Def of ≅
CPCTC
PS = RQ
m<3 = m <2
Given
Given
Def. of ≅
SAS
```