Let us join AC and PQ. ΔACQ and
ΔAQP are on the same base AQ and between the same parallels AQ and
CP.
∴ar(ΔACQ)=ar(ΔAPQ) ar(ΔACQ)−ar(ΔABQ)=ar(ΔAPQ)−ar(ΔABQ) [Subtracting ar
(ΔABQ) from both sides]
⇒ar(ΔABC)=ar(ΔQBP).......(1) Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively, ar(ΔABC)=12ar(ABCD)...(2) ar(ΔQBP)=12ar(PBQR)...(3) From equations (1), (2), and (3), we obtain; 12ar(ABCD)=12ar(PBQR) ⇒ar(ABCD)=ar(PBQR)