Join AC.
Now, $triangleADC and triangleADQ$ being on the same base AD and between the same parallels AD and BC, are equal in area.
$therefore ar(triangleADC)=ar(triangleADQ)$
$rArr ar(triangleADC)-ar(triangleADP)=ar(triangleADQ)-ar(triangleADP)$
$rArr ar(triangleAPC)=ar(triangleDPQ)." "...(i)$
Also, $triangle APC and triangleBPC$ being on the same base PC and between the same parallels PC and AB, are equal in area.
$therefore ar(triangleAPC)=ar(triangleBPC)." "...(ii)$
From (i) and (ii), we get $ar(triangleBPC)=ar(triangleDPQ)$.
