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)