In the adjoining figure, ABCD is a parallelogram and a line through A cuts DC at P and BC produced at Prove that

ar (∆BPC) = ar (∆DPQ)

Given ABCD is a parallelogram, so AB || CD and AD || BC. Lines AQ and DC intersect at P.

To prove ar (∆BPC) = ar (∆DPQ)

Construction Join AC.

As ∆ADC and ∆ADQ being on the same base AD and between the same parallel lines AD and BQ.

∴ ar (∆ADC) =ar (∆ADQ)

On subtracting ar (∆ADP) from both sides, we get

ar (∆ADC) - ar (∆ADP) = ar (∆ADQ) - ar (∆ADP)

ar (∆APC) = ar (∆DPQ)… (i)

Also, ∆APC and ∆BPC being on the same base PC and between the same parallel lines PC and AB.

So, they are equal in area,

i.e. ar (∆APC) =ar (∆BPC) …(ii)

From Eqs. (i) and (ii), we get

ar(∆BPC) = ar(∆DPQ)