TEQ * Typical Exam Questions

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TEQ – Typical Exam
Questions
TEQ #1
Statement
J
P
K
1
3
O
4
M
2
1. JKLM is a parallelogram
1. Given
2. JO  OL
2. Given
3. JPK MQL
3. Opposite sides of a
parallelogram are parallel
4. 1  2
4. Parallel lines cut by a
transversal form congruent
alternate interior angles
5. 3  4
5. Vertical angles are congruent
6. ΔJOP  ΔLOQ
6. ASA  ASA
7. OP  OQ
7. CPCTC
L
Q
Given: JKLM is a parallelogram
JO  OL
Prove:
OP  OQ
Reason
E
C
B
Statement
L
J
D
A
K
Given: Parallelogram DEBK,
, BC  DA
DJ  BL
Prove: CJ  AL
Reason
1. Parallelogram DEBK
1. Given
2. BC  DA
2. Given
3. DJ  BL
3. Given
4. JL  JL
4. Reflexive postulate
5. DJ  JL  BL  JL
5. Addition postulate
6. DJ  JL  DL
6. Partition postulate
BL  JL  JB
7. DL  JB
7. Substitution postulate
8. EB DK
8. Opposite sides of a parallelogram
are parallel
9. 1  2
9. Parallel lines cut by a
transversal form congruent
alternate interior angles.
10. ΔCJB  ΔALD
10. SAS  SAS
11. CJ  AL
11. CPCTC
TEQ #3
Statement
D
4
1
F
A
E
2
3
B
1. ABCD is a parallelogram
1. Given
2. DF  AC
2. Given
C 3. EB  AC
4. 1 and 2 are
right angles
Prove: DF  BE
3. Given
4. Perpendicular segments form
right angles
5. 1  2
5. All right angles are congruent
6. DC AB and DC  AB
6. Opposite sides of a
parallelogram are both parallel
and congruent
7. 3  4
7. Parallel lines cut by a
transversal form congruent
alternate interior angles
8. ΔDFC  ΔBEA
8. AAS  AAS
9. DF  BE
9. CPCTC
Given: Parallelogram ABCD,
DF  AC
Reason
TEQ #4
Statement
S
R
1
P
2
Q
Given : Parallelogram PQRS
PQ  QR
Prove: 1 is not congruent
to2
Reason
1. Parallelogram PQRS
1. Given
2. PQ  QR
2. Given
3. ∠1 ≅ ∠2
3. Assumption
4. PQRS is a rhombus
4. A parallelogram whose
diagonal bisects an angle is
a rhombus
5.  ≅ 
5. All sides of a rhombus
are congruent
6. ∠1   
∠2
6. Contradiction 2,5
TEQ #5
Statement
1. Rhombus ABCD
Reason
1. Given
2. E is the midpoint of  2. Given
4
1
3. DE  EF
3. A midpoint divides a segment
into two congruent parts
4. 1  2
4. Vertical angles are congruent
5. DC ABF
5. Opposite sides of a rhombus
are parallel
6. 3  4
6. Parallel lines cut by a
transversal form congruent
alternate interior angles.
7. ΔDEC  ΔFEB
7. ASA  ASA
8. DC  BF
8. CPCTC
9. DC  DA
9. All sides of a rhombus are
congruent
10. AD  BF
10. Substitution postulate
2
3
TEQ #6
Statement
Reason
1. ABDE is a parallelogram 1. Given
2. AE  DC
2. Given
3. AE  BD
3. Opposite sides of a
parallelogram are congruent
4. DC  BD
4. Substitution postulate
5. DBC is isosceles
5. A triangle with two congruent
sides is isosceles
Given: ABDE is a parallelogram
AE  DC
Prove : DBC is isosceles
TEQ #7
Statement
D
C
E
4
2
3
1
A
B
Reason
1. DB bisects AC
1. Given
2. 1  2
2. Given
3. AE  EC
3. A segment bisector divides a
segment into two congruent parts
4. 3  4
4. Vertical angles are congruent
5. ΔAED  ΔCEB
5. ASA  ASA
6. DA  BC
6. CPCTC
7. AD BC
7. Two lines cut by a transversal
that form congruent alternate
interior angles are parallel
8. ABCD is a parallelogram
8. A quadrilateral that has one
pair of opposite sides both
parallel and congruent is a
parallelogram
TEQ #9
Statement
Reason
1. AB  BC
1. Given
2. BD  BE
2. Given
3. A  C
3. Assumption
4 . B  B
4. Reflexive postulate
5. ABE  CBD
5. ASA  ASA
6. BD  BE
7. A  C
6. CPCTC
7. Contradiction 2,6
TEQ #10
Statement
D
A
E
F
C
B
1. ABCD is a parallelogram
1. Given
2. AE  FC
3. DA  CB
2. Given
Prove: ΔDAE  ΔBCF
3. Opposite sides of a
parallelogram are congruent
4. A  C
4. Opposite angles of a
parallelogram are congruent
5. ΔDAE  ΔBCF
5. SAS  SAS
Given: ABCD is a parallelogram
AE  FC
Reason
Statement
D
F
4
A
1
C
2
E
B
3
Reason
1. ABCD is a parallelogram
1. Given
2. DE  AC
3. BF  AC
2. Given
3. Given
4. 1 and 2 are
right angles
4. Perpendicular segments form
right angles
5. 1  2
5. All right angles are congruent
6. DA CB , DA  CB
6. Opposite sides of a
parallelogram are both congruent
and parallel.
7. 3  4
7. Parallel lines cut by a
transversal form congruent
alternate interior angles
8. ΔDEA  ΔBFC
8. AAS  AAS
9. AE  FC
9. CPCTC
Prove: AE  FC
TEQ #8
Q
P
Statement
T
R
Reason
1. PQ  RS
1. Given
2. PQ and RS intersect at T
2. Given
3. T is the midpoint of
PS and RQ
3. Assumption
4.  ≅ ,  ≅ 
4. A midpoint divides a
segment into two congruent
parts.
5. ∠1 ≅ ∠2
5. Vertical angles are
congruent
6. ∆ ≅ ∆
6.  ≅ 
S
Given : PQ  RS
7. ≅ 
PQ and RS intersect at T 8.T is not the midpoint
of PS and RQ
Prove : T is not the midpoint
of PS and RQ
7. CPCTC
8. Contradiction 1,7

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