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given : ABCD is parallelogram
DQ = QP = PB
TO PROVE :
CQ ║AP
PROOF:
In ΔADP and ΔCBQ
AD = BC [ opp. side of parallelogram]
DQ = BP - EQN 1
QP =QP - EQN 2
ADD EQN 1 AND EQN 2
DQ+QP = BP+QP
DP = BP
∠ADP =∠CBQ [ Alternate angle of parallelogram]
∴ΔADP ≅ ΔCBQ
∠APD = ∠BQC [CPCT]
If alternate angles are equal.
∴CQ ║AP
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