Answer Key to Review of Chapter 2
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Reflexive
Division
Addition
Reflexive
Transitive
Subtraction
Distributive
Division
Subtraction
Symmetric
Distributive
Substitution
See end
See end
1
4 x   6 Given
3
12 x  1  18 Mult.
12 x  17 Subtract.
17
Division
12
4( x  3)  23 Given
x
4 x  12  23 Distrib.
16. 4 x  11 Subtract.
11
x
Division
4
1
3x  5  x  7 Given
3
9 x  15  x  21 Multiplication
8 x  15  21 Subtraction
17.
8 x  6 Addition
4 x  3 Division
3
x
Division
4
AC  AB Given
AC  4 x  1 Given
AB  6 x  13 Given
18. 4 x  1  6 x  13 Substitution
2 x  1  13 Subtraction
2 x  14 Subtraction
x  7 Division
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
See End
See End
See End
See End
x  6, m5  40
mXWZ  106
x  30, m5  28
x  30, m7  50, m8  70, m9  60
x  58, m5  58
x  45, m1  86, m2  94
If angles are complementary, then their sum is
90. If 2 angles sum is 90, then they are
complementary. If angles are not
complementary, then their sum is not 90. If 2
angles sum is not 90, then they are not
complementary.
If you like hockey, then you go to the hockey
game. If you go to the hockey game, then you
like hockey. If you do not like hockey, then you
do not go to the hockey game. If you do not go
to the hockey game, then you do not like
hockey.
Law of Detachment
Invalid
Law of Syllogism
If a rectangle has 4 equal sides, then it is a
regular polygon. Law of Syllogism
A is a right angle. Law of Detachment
The correct conclusion should be “Angles C and
D are linear pair, so the angles are
supplementary.
13) Given: T is between Q and J. QT  KJ
Prove: KJ  TJ  QJ
T is between Q and J
J
T
Q
K
QT=KJ
Given
Given
QT+TJ=QJ
Seg. Add. Post.
KJ+TJ=QJ
Substitution Prop =
14) Given: M is the midpoint of YF
F is the midpoint of MD
Prove: 3YM YD
Y
M is the midpoint of YF
Given
F
M
D
F is the midpoint of MD
Given
MF=FD
YM=MF
Def. of Midpoint
Def. of Midpoint
YM=FD
YM+MF+FD=YD
Transitive Prop =
Seg. Add. Post.
YM+MF+YM=YD
Substitution Prop =
YM+YM+YM=YD
Substitution Prop =
3YM =YD
RD
RK
. .
Simplify
19) Given: RD  HK
Prove: RK  HD
R
KD
HK
Seg. Add. Post.
KD
D
DH
Seg. Add. Post.
Given
RD
RK
K
KD
HK
KD
DH
Substitution Prop =
RK
DH
Subtraction Prop =
H
... .
20) Given: mSWR  mHWF
Prove: mSWF  mRWH
S
F
R
H
W
mSWR  mSWF  mFWR
mFWH  mFWR  mRWH
mSWR  mHWF
Angle add. Post.
Angle add. Post.
Given
mSWF  mFWR  mFWR  mRWH
Substitution Prop =
mSWF  mRWH
Subtraction Prop. =
21) Given: line w bisects TB .
Prove: TA  AB .
.
w
A
T
B
Line w bisects TB
Given
A is the midpoint
Def. of Seg. Bisector
TA=AB
22) Given: DP  OP ,
Prove: GP is a segment bisector.
DP=OP
G
Def. of Midpoint
Given
P is the midpoint of DO
Def. of Midpoint
GP is a segment bisector
Def. of Seg. Bisector
D
P
O