Answer Key: Inequality and Indirect Test Review
1.
Statements
1)  YZX

(s)
2) m 1
m 2
(a )
3)

(s)
3) ZX > YW
Reasons
1) Given
2) Exterior POI
3) Reflexive Property
3) Hinge Theorem (1, 2, 3)
2.
1)
2)
3)
4)
Statements

(s)

(s)
DC < AB
(s )
m CBD m ADB
1)
2)
3)
4)
Statements


m 1
m 2
PS > QR
1)
2)
3)
4)
5)
6)
Statements
NK is a median of JMN
K is midpoint of JM

(s)
JN > NM
(s )

(s)
m 1
m 2
Reasons
1)
2)
3)
4)
Given
Reflexive Property
Given
Converse of the Hinge Theorem
(1, 2, 3)
3.
(s)
(s)
(a )
1)
2)
3)
4)
Reasons
Given
Reflexive Property
Exterior POI
Hinge Theorem (1, 3, 2)
4.
Reasons
1) Given
2) Def of median
3) Def of midpoint
4) Given
5) Reflexive Property
6) Converse of the Hinge Theorem
(3, 4, 5)
5.
Statements
1) AC < AE
2) m E < m C
3) D mp of
4)

5)

6) BD > FD
Reasons
1) Given
2)
(a )
3) Given
4) Def of midpoint
5) Given
6) Hinge Theorem (4, 2, 5)
(s)
(s)
6.
1)
2)
3)
4)
5)
6)
7)
Statements
T mp of ZX

(s)

(s)
SZ > WX (s )
m STZ
m WTX
STZ  XTR, WTX  ZTY
m XTR > m ZTY
Reasons
1) Given
2) Def of midpoint
3) Given
4) Given
5) Converse of the Hinge Theorem (2,3,4)
6) Vertical angles are congruent
7) Substitution POI (5, 6)
7.
Statements
1.


2. AB > DC
3.

4. AC > DB
5. m AFC > m DJB
(s) (s)
(s
)
Reasons
1. Given
2. Given
3. Reflexive Property
4. Addition POI (2, 3)
5. Converse of hinge theorem (1,1,4)
8.
1.
2.
3.
4.
5.
6.
Statements


K midpoint of

m SKL > m QKJ
LS > JQ
RS > QR
(s)
(s)
(a )
Reasons
1. Given
2. Given
3. Def of midpoint
4. Given
5. Hinge Theorem (1, 4, 3)
6. Addition POI (1, 5)
9.
Statements
1.

,
2. XUVW is a parallelogram
3. VW > XW
(s )
4.

(s)

(s)
6. m VZW > m XZW
7. XZU  VZW, UZV 
8. m XZU > m UZV
10.
Statements
1. Rectangle AFBC
2.

3. D midpoint of
4.

(s)
5.

6. CAD  ADC
7. m
> m CAD
8. m
> m ADC (a
9.

(s)
10. AE > AC
11. AE > FB
XZW
Reasons
1. Given
2. If rectangle → opposite sides
3. Given
4. Def of midpoint
5. Given
)
11.
Statements
1. BD bisects ABC
2. ABD  DBC
3.

, D not midpoint of
4. A  C
5. ABD  CBD
6.

7. D midpoint of AC
8 BD does not bisect <ABC
Reasons
1. Given
2. If a quad has 1 pr of sides both and ,
→
(1)
3. Given
4. If , → diagonals bisect each other.
5. Reflexive Property
6. Converse of the Hinge Theorem (3, 4, 5)
7. Vertical angles are congruent
8. Substitution POI (6, 7)
6.
7. Exterior POI
8. Substitution POI (6, 7)
9. Reflexive Property
10. Hinge Theorem (4, 8, 9)
11. Substitution POI (2, 10)
Reasons
1. Assumption Leading to a Contradiction
2. Def of angle bisector
3. Given
4.
5.
6.
7.
8.
ASA (2, 3, 4)
CPCTC
Def of midpoint
Contradiction (3, 7)
1.
2.
3.
4.
5.
6.
7.
8.
12.
Statements
 ADC is isosceles
ADB  CDB, ABC is not isosceles


ABD   CBD

ABC is isosceles
 ADC is not isosceles
Reasons
1. Assumption Leading to a Contradiction
2. Given
3. Definition of an Isosceles 
4. Reflexive Property
5. SAS (3, 2, 4)
6. CPCTC
7. Definition of an Isosceles 
8. Contradiction (2, 7)