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In Fig., X and Y are respectively the mid-points of the opposite sides AD and BC of a parallelogram ABCD. Also, BX and DY intersect AC at P and Q, respectively. Show that AP = PQ = QC.

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AD = BC (Opposite sides of a parallelogram)

Therefore, DX = BY ( 1/2 AD = 1/2 BC) 

Also, DX || BY (As AD || BC) 

So, XBYD is a parallelogram (A pair of opposite sides equal and parallel) 

i.e., PX || QD 

Therefore, AP = PQ (From ∆AQD where X is mid-point of AD) 

Similarly, from ∆CPB, CQ = PQ (1) 

Thus, AP = PQ = CQ [From (1) and (2)] (2)

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