Question

In the given figure ABCD is a parallelogram. If BC=CQ prove that ar(BCP)=ar(DPQ)

Answer 1

Explanation:

Given: ABCD is a parallelogram and BC=CQ.

To prove : ar(ΔBCP)=ar(ΔDPQ)

Construction : Join AC

Proof: ar(ΔBPC) = ar (ΔAPC) [Triangles on the same base and between same parallels] .Similarly ,

ar(ΔADC) = ar(ΔADQ).

Now, ar(ΔADC) = ar(ΔADP) +ar(ΔAPC) and

ar(ΔADQ) = ar(ΔADP) + ar(ΔDPQ) (from figure)

We know, ar(ΔADC) = ar(ΔADQ) and ar(ΔADP) is common.

Therefore, ar(ΔAPC) = ar(ΔDPQ)

But , ar(ΔAPC) = ar(ΔBPC)

Thus, ar(ΔBPC) = ar(ΔDPQ)

Hence proved.