In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure).
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Construction: Join AC to meet BD in O.
Therefore, OB=OD and OA=OC ...(1)
(Diagonals of a parallelogram bisect each other)
But BQ=DP ...given
∴OB–BQ=OD–DP
∴OQ=OP ....(2)
Now, in □APCQ,
OA=OC ....from (1)
OQ=OP ....from (2)
∴□APCQ is a parallelogram.
In △APD and △CQB,
AD=CB ....opposite sides of a parallelogram
AP=CQ ....opposite sides of a parallelogram
DP=BQ ...given
△APD≅△CQB ...By SSS test of congruence
∴AP=CQ ...c.s.c.t.
AQ=CP ...c.s.c.t. ...(3)
In △AQB and △CPD,
AB=CD ....opposite sides of a parallelogram
AQ=CP ...from (3)
BQ=DP ...given
∴△AQB≅△CPD ....By SSS test of congruence
Step-by-step explanation:
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