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202 Midterm Proof Review KEY
CHAPTER TWO PROOFS
1. Prove by algebraic proof: If
2 x  5
 5 , then x = −10
3
Statements
1.
−2𝑥−5
=
3
Reasons
5
1. Given
2. -2x-5 =15
3. -2x=20
4. x=-10
2. Multiplication/Substitution
3. Addition/Substitution
4. Division/Substitution
A
mABD  mBDF
m1  m4
Prove: 2  3
2. Given:
1
B
2
3
Statements
mABD  mBDF
1.
2.
3.
4.
5.
2.
3.
4.
5.
2. Angle Addition Postulate
3. Substitution
4. Subtraction
5. Def. of 
A
AC  DF
B C
BC  EF
Prove: AB  DE
Statements
1.
4
F
1. Given
m1  m4
mABD  m1  m2
mBDF  m3  m4
m1  m2  m3  m4
m2  m3
2  3
3. Given:
E
Reasons
AC  DF
BC  EF
AC=DF, BC = EF
AC = AB + BC, DF = DE + EF
AB + BC = DE + EF
AB = EF
6. AB  DE
D
E
Reasons
F
1. Given
2. Def. of 
3. Segment Addition Postulate
4. Substitution
5. Subtraction
6. Def. of 
C
D
4. Given: 2  3
Prove: 1  4
Statements
1
Reasons
1. 2  3
2. 1 and 2 are supplementary
3 and 4 are supplementary
3. 1  4
2
3
4
1. Given
2. Supplement Theorem
3. Angles Suppl. To Congruent Angles are Congruent
(Theorem 2.6)
S
O
P
H
5. Given: SO = PH
Prove: SP = OH
Statements
1.
2.
3.
4.
5.
Reasons
SO=PH
OP = OP
SO + OP = OP + PH
SO + OP = SP, OP + PH = OH
SP = OH
1. Given
2. Reflexive
3. Addition
4. Segment Addition Postulate
5. Substitution
6. Given: 𝑚∡𝐵𝐴𝐶 = 𝑚∡𝐹𝐸𝐺, 𝑚∡𝐶𝐴𝐷 = 𝑚∡𝐺𝐸𝐻
B
C
Prove: 𝑚∡𝐵𝐴𝐷 = 𝑚∡𝐹𝐸𝐻
Statements
Reasons
1. 𝑚∡𝐵𝐴𝐶 = 𝑚∡𝐹𝐸𝐺, 𝑚∡𝐶𝐴𝐷 = 𝑚∡𝐺𝐸𝐻
1. Given
mBAD  mBAC  mCAD
2. Angle Addition Postulate
mFED  mFEG  mGEH
3. mBAC  mCAD  mFEG  mGEH 3. Addition
4. 𝑚∡𝐵𝐴𝐷 = 𝑚∡𝐹𝐸𝐻
4. Substitution
D
A
2.
F
E
G
H
CHAPTER 3 PROOFS
̅̅̅̅ || 𝐶𝐷
̅̅̅̅
6. Given: 𝐴𝐵
CD bisects BCE
Prove: 1  3
A
B
1
Statements
̅̅̅̅
1. 𝐴𝐵 || ̅̅̅̅
𝐶𝐷, CD bisects BCE
2. ∡2 ≅ ∡3
3. ∡1 ≅ ∡2
4. 1  3
C
2 3
D
E
Reasons
1. Given
2. Angle Bisector Thm.
3. Alt. Int. Angle Thm.
4. Transitive (Subs)
7. Given: UD || HS
Statements
DH || US
1. Prove: 1  3
D
H
2
U
3
4
1
1. UD || HS, DH || US
2. 1 and 2 are supplementary
2 and 3 are supplementary
1. given
3. ∡2 ≅ ∡2
4. 1  3
3. Reflexive
4. Angles Suppl. To Congruent
Angles are Congruent (Thm 2.6)
Statements
a
b
1 2
4 3
5 6
8 7
9 10
12 11
m
2. Supplement Theorem
S
8. Given: a || b; l || m
Prove: 2 12
l
Reasons
1. a || b; l || m
1. given
2. ∡2 ≅ ∡6
3. ∡6 ≅ ∡12
4. 2 12
2. Corresp. Angles Post.
3. Alt. Ext. Angles Thm.
4. Transitive (Subs)
9. Given: 3 and 8 are supplementary
Prove: l || m
1
4
2
3
Reasons
Statements
Reasons
1. 3 and 8 are supplementary
2. 8 6
3. 3 and 6 are supplementary
4. l || m
l
6
5
m
8
1. Given
2. Vert. Angle Thm.
3. Substitution
4. Converse Consec. Int.
Angle Thm
7
CHAPTER 4 PROOFS
10. Given: 1 2, 5  6
Prove: WXY  WXZ
Statements
Y
Reasons
1 2, 5  6
2. 3 and 5 form a linear pair
4 and 6 form a linear pair
3. 3 and 5 are supplementary
4 and 6 are supplementary
1. Given
2. Def of linear pair
4. 3  4
4. Supplements of  ’s are 
5. WX  WX
6. WXY  WXZ
5. Reflexive Property
6. ASA
1.
W
1
2
3
4
3. Supplement theorem
Z
X5
6
11. Given: m1= m2
A
B
3
B is the midpoint of AC
BE
4
C
5
bi sec ts ABD
BD bi sec ts CBE
Prove: A  C
1
Statement
2
D
E
Reason
1. m1 = m2
2. BE  BD
1. Given
3. B is midpoint of AC
3. Given
4. AB  CB
5. BE bisects ABD
BD bisects CBE
4. Definition of midpoint
5. Given
6. m3 = m4
m4 = m5
6. Definition of  bisector
7. m3 =m5
8. ABE  CBD
9. A  C
7. Transitive Property
8. SAS
9. CPCTC
2. Conv. Isos.  Theorem
̅̅̅̅ ≅ ̅̅̅̅
Given:𝐵𝐷
𝐶𝐸 ; ̅̅̅̅
𝐶𝐺 ≅ ̅̅̅̅
𝐷𝐺 ; ∡𝐴 ≅ ∡𝐹
Prove: AB  FE
Statement
F
A
12.
G
Reason
B
1. BC = DE; m1 = m4; AD  FC
2. 1 and 2 form a linear pair
3 and 4 form a linear pair
3. 1 and 2 are supplementary
3 and 4 are supplementary
4. 2  3
5. CD = CD
1. Given
2. Def of linear pair
6. BC + CD = DE + CD
6. Addition
6. BC + CD = BD
DE + CD = CE
7. BD = CE
7. Segment addition postulate
9. ABD  FEC
10. AB  FE
9. SAS
10. CPCTC
3. Supplement theorem
4. Supplements of  ’s are 
5. Reflexive
8. Substitution
1
C
2
3
4
D
E
13. Given: SO bisects TOP , PO  OT
Prove: PS  TS
Statements
1. SO bisects TOP , PO  OT
2. ∡𝟑 ≅ ∡𝟒
3. ̅̅̅̅
𝑺𝑶 ≅ ̅̅̅̅
𝑺𝑶
4. ∆𝑷𝑺𝑶 ≅ ∆𝑻𝑺𝑶
5. PS  TS
P
S
Reasons
1. Given
2. Angle Bisector Thm.
3. Reflexive
4. SAS
5. CPCTC
1
2
3
4
O
T
H
T
5
14. Given: 1  2 3, ES  DT
Prove: 4  5
1
2
E
1.
2.
3.
4.
Statement
1  2  3, ES  DT
HE  HD
HDT  HES
4  5
1.
2.
3.
4.
3
4
S
D
Reason
Given
Conv. Isosceles  Theorem
SAS
CPCTC
B
A
1
3
2
4
15. Given: ∡5
≅ ∡6 , ∡7 ≅ ∡8
Prove: AB  AD
Statements
1. ∡5 ≅ ∡6, ∡7 ≅ ∡8
̅̅̅̅ ≅ 𝐸𝐶
̅̅̅̅
2. 𝐸𝐶
3. ∆𝐵𝐸𝐶 ≅ ∆𝐷𝐸𝐶
̅̅̅̅ ≅ 𝐴𝐶
̅̅̅̅
4. 𝐴𝐶
̅̅̅̅
5. 𝐵𝐶 ≅ ̅̅̅̅
𝐷𝐶
6. ∆𝐴𝐵𝐶 ≅ ∆𝐴𝐷𝐶
7. ̅̅̅̅
𝐴𝐵 ≅ ̅̅̅̅
𝐴𝐷
7
8
E
5
6
C
D
Reasons
1. Given
2. Reflexive
3. ASA
4. Reflexive
5. CPCTC
3. SAS
7. CPCTC
D
̅̅̅̅  𝐶𝐸
̅̅̅̅
16. Given: 1  2 , 𝐴𝐵
Prove: BD  ED
Statements
1. 1  2 , ̅̅̅̅
𝐴𝐵  ̅̅̅̅
𝐶𝐸
̅̅̅̅ ≅ 𝐷𝐶
̅̅̅̅
2. 𝐴𝐷
3. ∆𝐴𝐷𝐵 ≅ ∆𝐶𝐷𝐸
4. BD  ED
Reasons
A
1
B
1. Given
2. Converse Isos. Triangle Thm.
3. SAS
4. CPCTC
E
2
C
ANSWER KEY Geometry 202 Review of Coordinate Proofs
1. Given AC = DF find the remaining coordinates of points B and E.
y
C( 2v , 2t)
B( v, t )
A( 0, 0)
D ( 3v , 2t )
E ( 4v , t)
F (5v , 0 )
x
B(v , t )
E(4v, t )
a) Prove BE|| CD || AF
Slope of BE =
t t
0
2t  2t 0
00
0

 0 . Slope of CD =
  0 . Slope of AF =

0
4v  v 3v
3v  2v v
5v  0 5v
Therefore, since all three slopes are equal, by definition of parallel lines, BE || CD || AF.
b) Prove BE = ½ (CD + AF)
Since CE, CD, and AF are horizontal lines, the distance can be found by d=| x2  x1 |
BE = |4v-v| = 3v
CD = |3v-2v| = v
CD + AF = v + 5v = 6v.
AF = | 5v-0| = 5v
½(6v) = 3v
Therefore BE = ½ (CD + AF)
BE  (4v  v ) 2  (t  t ) 2
CD  (3v  2v ) 2  (2t  2t ) 2
AF  (5v  0) 2  (0  0) 2
If you prefer, you can use the distance formula.
2. Use coordinate geometry to prove that the medians of an equilateral triangle are congruent. Complete the
given diagram to find missing points.
y
H(2a, 2a
J(? , ?)
G(0,0)
O
)3
K(? , ?)
I(4a, 0)
L(? , ?)
x
J(a, a 3 )
K( 3a , a 3 )
L( 2a, 0)
Now prove:
GK  IJ  HL :
GK = (3a  0)2  (a 3  0)2 = 9a2  3a2  12a2  2a 3
IJ = (4a  a)2  (0  a 3 )2  (3a)2  (a 3 )2  9a2  3a2  12a2  2a 3
HL = (2a  2a)2  (0  2a 3)2  0  (2a 3 )2  4a2 (3)  2a 3
Therefore GK = IJ = HL, by definition of congruence, their segments are congruent.