1. Prove that in a parallelogram, the diagonals bisect
each other.
Given: Parallelogram ABCD with diagonals
meeting at F.
Prove: AF = FC and BF = FD
A
Statement
Reason
1. AD = BC
Parallelogram property.
2.  BAC =  DCA
3.  ABD =  CDB
4. ∆ ABF  ∆CDF
Parallelogram property.
Parallelogram property.
5. AF = FC
6. BF = FD
QED
ASA
CPCTE
CPCTE
§ 7.1
D
C
F
B
2. Prove that a quadrilateral in which both
pairs of opposite sides are congruent is a
parallelogram.
D
C
A
B
Given: Quadrilateral with AB = CD and BC = AD.
Prove: Parallelogram
Statement
Reason
1. AC is diagonal
Construction.
2. AB = CD and BC = AD
3. ∆ ABC  ∆CDA
Given
SSS
4.  D =  B
5. ABCD parallelogram
CPCTE
Opposite angles
congruent
QED
D
3. Prove that if the diagonals of a quadrilateral
bisect each other, then it is a parallelogram.
Given: Quadrilateral with AF = FC and BF = FD.
Prove: Parallelogram
Statement
Reason
1. AF = FC AND BF = FD
Given
2.  AFD =  CFB
3. ∆ AFD  ∆CFB
Vertical angles
SAS
4. AD = BC
5. ∆ AFB  ∆CFD
CPCTE
6. AB = CD
7. ∆ AFB  ∆CFD
8. ABCD is a parallelogram.
QED
C
F
A
SAS
CPCTE
SAS
Opposite sides equal.
B
D
4. Prove that in a rhombus the diagonals are
perpendicular to one another.
E
A
Given: Rhombus with diagonals
Statement
Reason
1. Diagonals bisect one another
True of parallograms
2. 4 interior ∆s are all congruent
3. ∆ ABC  ∆CDA
SSS
4. 4 angles around E are all equal
and thus 90º
5. AC  BD
QED
SSS
Thm.
Def 
C
B
5 Midpoint Connector
Given: ΔABC, L and M midpoints.
Prove: LM ∥ BC and
LM = ½ BC
(1) Extend LM to P so that LM = MP
Construction
SAS
Alt inter s, A & ACP
Construction
Alt inter s
SAS
CPCTE
Alt inter s, BCL & PLC
(2)  ALM ≅  CPM
(3) AB ∥ PC
(4) Construct LC
(5) BLC ≅ PCL
(6)  LCP ≅  CLB
(7) LM = ½ LP = ½ BC
(8) LM ∥ BC
A
L
B
M
P
C
5
6. Prove that the perpendicular bisectors of the two diagonals and the
four sides of a cyclic quadrilateral are concurrent.
This one is easy. The four sides and the diagonals are all chords and the
perpendicular bisectors of all chords go through the center of the circle.
7. Prove that the opposite angles of a cyclic quadrilateral are
supplementary.
This takes some knowledge of high school geometry and inscribed
angles. An inscribed angle is equal in measure to ½ the intercepted
arc. The two opposite angles intercept arcs that total the measure of
the entire circumference of the circle. Thus the measure of the two
opposite angles is ½ of 360 or 180 and thus they are cupplementary.