ABCD is a parallelogram. BC is produced to Q such that BC = CQ. Then

Join AC

$$\because$$ $$\triangle$$APC and $$\triangle$$BPC lie on the same base CP and between same parallels AB & PC.

=> ar($$\triangle$$APC) = ar($$\triangle$$)BPC

Now, AD | | CQ and AD = CQ

=> ADQC is a parallelogram

Similarly, ar($$\triangle$$ADC) = ar($$\triangle$$ADQ)

Subtracting ar($$\triangle$$ADP) from both sides, we get :

=> ar($$\triangle$$APC) = ar($$\triangle$$DPQ)

$$\therefore$$ ar($$\triangle$$BPC) = ar($$\triangle$$DPQ)