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Geometry Ch 2 Proof Practice
1.
Name_____________________________
Per_______
W
Y
X

Given: m  1 = m  3
Prove: m  WOY = m  ZOX
1
2
3
Statements
Reasons
O
1) m  1 = m  3
1)
Given
2) m  2 = m  2
2)
Reflexive
3) m  1 + m  2 = m  3 + m  2
3)
Addition prop. of =
4) m  WOY = m  1 + m  2
4)
Angle Addition Postulate
5)
Substitution

m  ZOX = m  3 + m  2
5) m  WOY = m  ZOX
2.
Given: LH = AS
Prove: LA = HS
L
●
●
Statements
S
H
A
●
●
Reasons
1) LH = AS
1)
Given
2) LH = LA + AH
2)
Segment Addition Postulate
3) LA + AH = AH + HS
3)
Substitution prop.
4) AH = AH
4)
Reflexive prop. of =
5) LA = HS
5)
Subtraction prop. of =
AS = AH + HS
3.
Given: WX = YZ; Y is the midpoint of XZ
Prove: WX = XY
W
X
●
Statements
●
Y
●
Reasons
1) WX = YZ; Y is the midpoint of XZ
1)
Given
2) XY = YZ
2)
Def. of Midpoint
3) WX = XY
3)
Substitution / Transitive prop. of =
Z
●
Z
N
●
●I
4.
Given: FR bisects IFG
Prove: 2  3
●
G
Statements
F
●
2
1
R●
3
●
S
Reasons
1) FR bisects IFG
1)
Given
2) ∠1
∠2
2)
Def. of Angle Bisector
3) ∠1
∠3
3)
Vertical Angles are Congruent
(Vertical Angles Theorem)
4) ∠2
∠3
4)
Substitution / Transitive prop. of
5.
●D
Given: AB  DC
m1  m2
Prove: m2  90
●
A
●1
C
Statements
●
B
2
Reasons
1) AB  DC ; m1  m2
1)
Given
2) m1  90
2)
Def. of Perpendicular Lines
3) m2  90
3)
Substitution / Transitive prop. of =
6.
Given: m1  m3  180
Prove: m2  m3
1
Statements
2
3
Reasons
1) m1  m3  180
1)
Given
2) m1  m2  180
2)
Def. of Supplementary Angles / Supp. Thm
3) m1  m2  m1  m3
3)
Substitution
4) m1  m1
4)
Reflexive
5) m2  m3
5)
Subtraction
6) 2  3
6)
Def of cong.