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CP1 Math 2
December Cumulative Review SOLUTIONS
1. Given: ̅̅̅̅
𝐸𝐴 ∥ ̅̅̅̅
𝐷𝐵
̅̅̅̅
𝐸𝐴 ≅ ̅̅̅̅
𝐷𝐵
̅̅̅̅
B is the midpoint of 𝐴𝐶
̅̅̅̅
̅̅̅̅ ∥ 𝐷𝐶
Prove: 𝐸𝐵
Statement
1. ̅̅̅̅
𝐸𝐴 ∥ ̅̅̅̅
𝐷𝐵
2. ∠𝐸𝐴𝐵 ≅ ∠𝐷𝐵𝐶
3. ̅̅̅̅
𝐸𝐴 ≅ ̅̅̅̅
𝐷𝐵
4. B is the midpoint of ̅̅̅̅
𝐴𝐶
5. 𝐴𝐵 ≅ 𝐵𝐶
6. ∆𝐸𝐴𝐵 ≅ ∆𝐷𝐵𝐶
7. ∠𝐸𝐵𝐴 ≅ ∠𝐷𝐶𝐵
8. ̅̅̅̅
𝐸𝐵 ∥ ̅̅̅̅
𝐷𝐶
Reason
1. Given
2. Corresponding Angles Theorem
3. Given
4. Given
5. Definition of midpoint
6. SAS (steps 3, 2, 5)
7. CPCTC
8. Converse of corresponding angles theorem
̅̅̅̅ ≅ ̅̅̅̅
2. Given: 𝐴𝐹
𝐴𝐸
̅̅̅̅ ≅ ̅̅̅̅
𝐹𝐷
𝐸𝐷
∠1 ≅ ∠2
̅̅̅̅
Prove: ̅̅̅̅
𝐴𝐷 ⊥ 𝐶𝐵
Statement
̅̅̅̅ ≅ ̅̅̅̅
1. 𝐴𝐹
𝐴𝐸
̅̅̅̅ ≅ ̅̅̅̅
2. 𝐹𝐷
𝐸𝐷
3. ̅̅̅̅
𝐴𝐷 ≅ ̅̅̅̅
𝐴𝐷
4. ∆𝐸𝐴𝐷 ≅ ∆𝐹𝐴𝐷
5. ∠𝐸𝐷𝐴 ≅ ∠𝐹𝐷𝐴
6. ∠1 ≅ ∠2
7. ∠𝐶𝐷𝐴 = ∠1 + ∠𝐸𝐷𝐴 and ∠𝐵𝐷𝐴 = ∠2 +
∠𝐹𝐷𝐴
8. ∠𝐶𝐷𝐴 ≅ ∠𝐵𝐷𝐴
9. ∠𝐶𝐷𝐴 and ∠𝐵𝐷𝐴 are supplementary
10. ∠𝐵𝐷𝐴 = ∠𝐶𝐷𝐴 = 90°
̅̅̅̅ ⊥ 𝐶𝐵
̅̅̅̅
11. 𝐴𝐷
Reason
1. Given
2. Given
3. Reflexive Property
4. SSS
5. CPCTC
6. Given
7. Whole = sum of parts
8. Addition property
9. They form a straight line
10. They are supplementary and congruent
11. Definition of perpendicular
K
3. Given: ∠1 ≅ ∠2
𝐹𝐺 ≅ 𝐾𝑀
𝐺𝐽 ≅ 𝐻𝐾
Prove: 𝐹𝐻 ∥ 𝐽𝑀
Statement
1. 𝐺𝐽 ≅ 𝐻𝐾
2. 𝐺𝐽 = 𝐺𝐻 + 𝐻𝐽 and 𝐻𝐾 = 𝐽𝐾 + 𝐻𝐽
3. 𝐺𝐻 + 𝐻𝐽 ≅ 𝐽𝐾 + 𝐻𝐽
4. 𝐺𝐻 ≅ 𝐽𝐾
5. 𝐹𝐺 ≅ 𝐾𝑀
6. ∠1 ≅ ∠2
7. ∆𝐹𝐺𝐻 ≅ ∆𝑀𝐾𝐽
8. ∠3 ≅ ∠4
9. 𝐹𝐻 ∥ 𝐽𝑀
2
F
J
3
4
H
1
G
Reason
1. Given
2. Segment addition (whole = sum of parts)
3. Substitution property
4. Subtraction property
5. Given
6. Given
7. SAS
8. CPCTC
9. Converse of Alt. Ext. Angles Theorem
̅̅̅̅
4. Given: ̅̅̅̅
𝐴𝐵 ≅ 𝐵𝐶
̅̅̅̅
𝐴𝐷 ≅ ̅̅̅̅
𝐷𝐶
Prove: ∠1 ≅ ∠2
Statement
̅̅̅̅ ≅ 𝐷𝐶
̅̅̅̅
1. 𝐴𝐷
2.
3.
4.
5.
6.
7.
∠3 ≅ ∠4
̅̅̅̅
̅̅̅̅
𝐴𝐵 ≅ 𝐵𝐶
∠𝐵𝐴𝐶 ≅ ∠𝐵𝐶𝐴
∠𝐵𝐴𝐶 = ∠1 + ∠3 and ∠𝐵𝐶𝐴 = ∠2 + ∠4
∠1 + ∠3 ≅ ∠2 + ∠4
∠1 ≅ ∠2
Reason
1. Given
2. ITT
3. Given
4. ITT
5. Angle Addition (whole = sum of parts)
6. Substitution
7. Subtraction Property
M
5. A twist on what you have done before …
Given: Transversal t cuts a and b.
∠1 ≇ ∠2
Prove: ∠1 ≇ ∠3
6.
Statement
Reason
1. ∠1 ≇ ∠2
1. Given
2. ∠2 ≅ ∠3
2. VAT (Vertical angles are congruent)
3. ∠1 ≇ ∠3
3. Transitive property
Given: ̅̅̅̅
𝐶𝐹 ≅ ̅̅̅̅
𝐶𝐸 ≅ ̅̅̅̅
𝐸𝐷
Show that 𝑦 = 3𝑥.
Algebra:
Reasons (but you don’t need these as this is not a proof)
∠𝐷 = 𝑥
∠𝐶𝐸𝐷 = 180 − 2𝑥
∠𝐶𝐸𝐹 = 2𝑥
∠𝐶𝐹𝐸 = 2𝑥
∠𝐹𝐶𝐸 = 180 − 4𝑥
∠𝐴𝐶𝐷 = 180 = 𝑦 + (180 − 4𝑥) + 𝑥
0 = 𝑦 − 3𝑥
3𝑥 = 𝑦
(ITT)
(Triangle Angle Sum)
(Supplementary to ∠𝐶𝐸𝐷)
(ITT)
(Triangle Angle Sum)
(Definition of straight angle)
(Algebra)
7. Given: 𝐴𝐵𝐶𝐷 and 𝐴𝐵𝑋𝑌 are parallelograms
̅̅̅̅̅
̅̅̅̅̅ ≅ 𝑀𝐶
𝐵𝑀
𝐷𝐶𝑌𝑋 is a straight line
Prove: Points C and Y trisect 𝐷𝐶𝑌𝑋
Statement
1. 𝐴𝐵𝐶𝐷 and 𝐴𝐵𝑋𝑌 are parallelograms
2. 𝐴𝐵 ≅ 𝐷𝐶 and 𝐴𝐵 ≅ 𝑋𝑌
3. 𝐷𝑋 ∥ 𝐴𝐵
4. ∠𝑌𝐶𝑀 ≅ ∠𝐴𝐵𝑀
5. ∠𝐶𝑀𝑌 ≅ ∠𝐵𝑀𝐴
6. Δ𝐶𝑀𝑌 ≅ ∆𝐵𝑀𝐴
7. 𝐴𝐵 ≅ 𝐶𝑌
8. 𝐷𝐶 ≅ 𝐶𝑌 ≅ 𝑋𝑌
9. Points C and Y trisect 𝐷𝐶𝑌𝑋
Reason
1. Given
2. Opposite sides of parallelograms are congruent
3. Definition of parallelogram
4. PAI
5. VAT (Vertical angles are congruent)
6. ASA
7. CPCTC
8. Transitive Property (they are all congruent to 𝐴𝐵)
9. Definition of trisect
8. Given: 𝐸𝐷 ≅ 𝐸𝐹
𝐸𝐻 ≅ 𝐸𝐺
∠1 ≅ ∠2
Prove: ∆𝐷𝐽𝐻 ≅ ∆𝐹𝐾𝐺
Statement
1. 𝐸𝐷 ≅ 𝐸𝐹
2. ∠5 ≅ ∠6
3.
4.
5.
6.
7.
8.
9.
𝐸𝐻 ≅ 𝐸𝐺
∠1 ≅ ∠2
Δ𝐸𝐷𝐻 ≅ ∆𝐸𝐹𝐺
∠𝐸𝐻𝐷 ≅ ∠𝐸𝐺𝐹
𝐷𝐻 ≅ 𝐺𝐹
∠𝐸𝐷𝐻 ≅ ∠𝐸𝐹𝐺
∠𝐸𝐷𝐻 = ∠5 + ∠3 and ∠𝐸𝐹𝐺 = ∠6 + ∠4
Reason
1. Given
2. ITT
3. Given
4.
5.
6.
7.
8.
9.
Given
SAS (steps 1, 4, 3)
CPCTC
CPCTC
CPCTC
Angle addition (whole = sum of parts)
10. ∠3 ≅ ∠4
10. Addition property
11. ∆𝐷𝐽𝐻 ≅ ∆𝐹𝐾𝐺
11. AAS
9. True
10. False. 𝐴𝐵𝐶𝐷 could be a rectangle or an isosceles trapezoid.
11. False. 𝐴𝐵𝐶𝐷 could be a kite. But if BOTH pairs of opposite angles were congruent, then this statement
would be true.
12. True. You can prove this if you want extra proof-writing practice 
13. Given 𝑁𝐼𝐶𝐸 is a kite. 𝑁𝐸 = √6;
𝑀𝐼 = 2; 𝐸𝐶 = 2√6
a. Find the perimeter of 𝑁𝐼𝐶𝐸.
√6 + √6 + 2√6 + 2√6 = 𝟔√𝟔
b. Find the area of 𝑁𝐼𝐶𝐸.
First use Pythagorean Theorem to find the lengths of 𝑁𝑀 and 𝑀𝐶:
(√6)2 = 22 + 𝑁𝑀
2
(2√6)2 = 22 + 𝑀𝐶
2
2
2
𝑁𝑀 = 2
𝑁𝑀 = √2
𝑀𝐶 = 20
𝑀𝐶 = √20 = 2√5
1
Area of ∆𝐸𝐶𝐼 = 2 (4)(2√5) = 4√5
1
Area of ∆𝐸𝑁𝐼 = 2 (4)(√2) = 2√2
Total area = 𝟒√𝟓 + 𝟐√𝟐
(Work shown for #14-16 on the pages following)
14. a) −64
1
c) 2
15. a)
2√10
25
16. a) 𝑥 = 1
b) No real solutions because you cannot take an even root of a negative number
d)
𝑦8
25
9
6
1
f) 2𝑦 2 ∙ ( 3 )
e) 4𝑥 4
2
b) 25 + 5√5
b) 𝑥 = 2
√𝑥
c)
3
c) 𝑥 = 4
√7
7
𝑏
4
g) 𝑎 ∙ √4
Work for #14-16: