Given
Segment Addition Postulate
Defn of Segment Bisector
Defn of Midpoint
Defn of
Defn of Right Angles
Defn of Perpendicular Lines
Angle Addition Postulate
Defn of Angle Bisector
Defn of Complementary Angles
Defn of Supplementary Angles
Angles Supp to Same Angle are
Angles Comp to Same Angle
Definition of a Linear Pair
Linear Pairs are Supp Angles
Vertical Angles are
Angles
Add Property
Subtraction Property
1. Given: PR = QS
Prove : PQ = RS
P
Q
Statement
R
Mult Property
Divide Property
Distribute Property
Reflexive Property
Symmetric Property
Transitive Property
Substitution Property
Simplify
S
Reasons
1. PR = QS
1.
2. PQ + QR = PR
2.
3. QR + RS = QS
3.
4. PQ + QR = QR + RS
4.
5. PQ = RS
5.
2. Given : m
AXC = m
DYF, m
1=m
3
2=m
C
Statement
AXC = m DYF,
1.
1=m
3. m
AXC = m
1+m
2
3.
4. m
DYF = m
3+m
4
4.
5. m
1+m
2=m
3+m
4
5.
6. m
3+m
2=m
3+m
4
6.
7. m
2=m
4
3
2.
7.
Q
P
3. Given : m
1= m 2
Prove: m PXR = m SXQ
2
1. m
Statement
1= m 2
1.
2. m
3= m
3
2.
3. m
1+ m
3=m
4.
PXR =m
1+m
3
4.
SXQ = m
2+ m
3
5.
6. m
PXR = m
SXQ
R
3
1
X
5. m
4
Y
Reasons
2. m
m
3
4
X
1. m
F
D
2
1
Prove : m
E
B
A
S
Reasons
2+ m
3
3.
6.
C
A
B
D
4. Given : AB = CD , BD = DE
Prove : AD = CE
E
Statement
Reasons
1.
1. Given
2. Given
2.
3. Segment Addition Postulate
3.
4. Segment Addition Postulate
4.
5. Substitution Property
5.
6. Substitution Property
6.
5. Given: AD = 2AB + BC
Prove: AB = CD
A
B
Statement
C
Reasons
1.
1. Given
2.
2.
3.
3. Segment Addition Postulate
4.
4. Substitution Property
5.
5. Substitution Property
6.
6. Subtraction Property
7.
7. Subtraction Property
Segment Addition Postulate
D
6.
Given: m
Prove : m
AEC = m
AEB = m
DEB
DEC
B
A
C
E
D
Statement
Reasons
1.
1. Given
2.
2. Angle Addition Postulate
3.
3. Angle Addition Postulate
4.
4. Substitution
5.
5. Subtraction
T
7. Given:
V
S
bisects
m
2=m
YXV,
3
1= m
3
1
R
Prove: m
W
2
4
X
4
Statement
Reasons
1.
1. Given
2.
2. Definition of angle bisector
3.
3. Given
4.
4. Definition of an angle bisector
5.
5. Given
Y
6.
6. Substitution Property
7.
7. Substitution Property
M
L
8. Given: m
LON = m
2
3
O
1=
C
3
Statement
Reasons
1.
1. Given
2.
2.
3.
3. Definition
4.
4.
5.
5. Definition
6. m
LON = m
1+
of bisector
of bisector
6.
2
7.
7.Angle Addition postulate
8.
8.Substiution property
9. m
1+
1=m
3+
4
9.
10. m
1+
1=m
3+
3
10.
11. 2m
12. m
1= 2 m
1= m
D
DCF
1
Prove:
E
F
N
3
3
11.
12. Division property
4
C
B
9. Given:
m
BAC + m
Prove: m
BAC =m
D
DAT = 90
A
CAD
T
Statement
1.
1. Given
2.
2. Given
3.
3. Def. of Perpendicular Lines
4.
4. Angle Addition Postulate
5. m
CAT = 90
5.
6.
6.Substitution
7.
7.Substitution
8. m
10. Given:
Prove:
Reasons
7
1
CAD = m
BAC
8.
6
4
1
8
Statement
2
7
3 4
6
5
Reasons
D
M
A
B
11. Given: M is the midpoint of
X is the midpoint of
AM = CX
X
C
Prove: MB = XD
Statement
Reasons
3
12. Given:
1 and
2 form a linear pair
m 2 + m 3 +m 4 =180
Prove: m 1=m 3 +m 4
1
Statement
2
4
Reasons
13.
Given: RS
TV
Prove: RT
SV
R
S
V
T
Statement
Reasons
D
14. Given:
, AC = KL
A
Prove: CB + LT = KT
K
Statement
C
L
Reasons
B
T