Question

ABCD is a parallelogram in which p is the mid point of DC.A line through C drawn paralell to PA meets DA produced at Q and AB at R.prove that DQ = 2AD and CQ = 2CR.

Answer 1

In this question make a figure like this made above.
Here,
∆ADP will be similar to ∆CDQ
[Angle D is common
AP is parallel to CQ
so angle A is equal to angle Q
so here by AA, triangles are similar]
By BPT theorem,
AD/AQ = DP/PC
Here, DP=PC
So, in LHS,
AD = AQ
Hence,proved in 1st part
then, AR is parallel to CD
so ∆ARQ is similar to ∆DCQ
By BPT theorem,
AQ/AD = QR/RC
AQ = AD. [Proved above]
so,
QR = RC
Hence, Proved.
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