2.5 Proofs for Segment and Angle Relationships
Name _________________________
Fill in the missing statement or reasons for the following proofs.
#1 Supplementary Angles
Given: <1 and <2 are supplementary
o
m<1 = 68
1
2
o
Prove: m<2 = 112
Statement
1.
2.
3.
4.
5.
Reason
1. Given
2.
3.
4. Substitution Property
5.
m∠1 + m∠2 =180
m∠1 =68
m∠2 =
112
#2 Segment Midpoint
Draw your own diagram and label
Given: M is the midpoint of LN ; LM = 8
Prove: LN = 16
Statement
1. M is the midpoint of LN ; LM = 8
2.
3.
4. MN = 8
5. LN
= LM + MN
6.
#3 Vertical Angles
Given: ∠1 ≅ ∠2, m∠4 =30

Reason
1.
2. Defn. of a midpoint
3. Defn. of congruent segments
4.
5.
6. Substitution
2
1
3
6
4
5
Prove: m∠2 =
30
Statement
1.
2. ∠1 ≅ ∠4
3.
4. m∠2 = m∠4
5. m∠4 =30
6.
Reason
1. Given
2.
3. Transitive Property
4. Defn. of congruent angles
5.
6.
A
#4 Complementary Angles

Given: <1 is complementary to <2; BE bisects ∠DBC
D
E
Prove: <1 is complementary to <3
B
Statement

1. BE bisects ∠DBC
2. ∠2 ≅ ∠3
3.
4. <1 is complementary to <2
5. m∠1 + m∠2 =______ 
6. m∠1 + m∠3 =90
7.
C
Reason
1.
2.
3. Defn. of congruent angles
4.
5.
6.
7. Defn. of complementary angles
#5 Segment Addition
Given: AB =+
3 x 5; BC =
2( AB ); AC =
60
Statement
1. AB =+
3 x 5; BC =
2( AB ); AC =
60
2. =
BC 2(3 x + 5)
3.
4. 3 x + 5 + 6 x + 10 =
60
5.
6. 9 x = 45
7.
8.=
BC 2 [ 3(5) + 5]
9.
C
B
A
Prove: BC = 40
Reason
1.
2.
3.
4.
5.
6.
7.
8.
9.
Segment addition postulate
Simplify
Division property
Simplify
#6 Segment Addition
X
Given:=
XY 6,=
XZ 14
Y
Z
Prove: YZ = 8
Statement
1.
2.
3.
4.
=
XY 6,=
XZ 14
XY + YZ =
XZ
6 + YZ =
14
YZ = 8
Reason
1.
2.
3.
4.
Q
P
#7 Midpoint of a Segment
R
Given: Q is the midpoint of PR ; QR ≅ RS
Prove: PQ ≅ RS
S
Statement
1. Q is the midpoint of PR ; QR ≅ RS
2. PQ ≅ QR
3. PQ ≅ RS
Reason
1.
2.
3.
#8 Supplementary Angles
Given: <1 and <2 are supplementary; m∠2= 4 x + 30 ; m∠1 =x
2
Prove: m∠2 =
150
Statement
Reason
1. <1 and <2 are supplementary;
1.
2.
3.
4.
5.
6.
7.
2.
3.
4.
5.
6.
7.
m∠2= 4 x + 30 ; m∠1 =x
m∠1 + m∠2 =180
x + 4 x + 30 =
180
5 x + 30 =
180
5 x = 150
x = 30
m∠
=
2 4(30) + =
30 150 =
A
#9 Angle Bisector

Given: BF bisects ∠ABC
5
B
Prove: ∠1 ≅ ∠6
1
Statement
1.
2.
3.
4.

BF bisects ∠ABC
∠5 ≅ ∠6
∠1 ≅ ∠5
∠1 ≅ ∠6
F
6
C
Reason
1.
2.
3.
4.
1
#10 Angle Addition Postulate
E
G
o
Given: ∠GFI =
89
(10x - 7) 2xo
Prove: x = 8
F
Statement
#11 Vertical Angles
I
Reason
2
1
Given: ∠1 ≅ ∠2
3
6
4
5
Prove: ∠4 ≅ ∠5
Statement
Reason
#12 Supplementary Angles
2
Given: <1 and <2 are supplementary; m∠1= 120 − x ; m∠2 =
2x
Prove: x = 60
Statement
Reason
1
#13 Segment Addition
Given: AC = 8 x , AB
= 2 x + 1 , BC
= 4x + 3
C
B
A
Prove: x = 2
Statement
#14 Bisecting Angles

Given: BE bisects ∠AEC
Prove: x = 2
Reason
A
B
6x - 1 10x - 9
E
Statement
Reason
#15 Midpoints on segments
Given: Point I is the midpoint of BG , GI = 25 , BI
= 7x + 4
Prove: x = 3
Statement
Reason
C