In figure X and Y are the mid-points of AC and AB respectively, `QP || BC` and CYQ and BXP are straight lines. Prove that `ar (DeltaABP) = ar (DeltaACQ)`.

Explanation

In `trianglePAC`, we have PA||DE and E is the midpoint of AC.
`therefore` D is the midpoint of PC (by converse of midpoint theorem).
And so, `DE = (1)/(2)PA`, i.e., `PA = 2(DE)`.
Similarly, AQ = 2(DE) and so, `PA = AQ`.
Now, `ar(triangleABQ)=ar(triangleACP)`
[`therefore triangleABQ and triangleACP` have equal base and lie between the same parallels].