In the following figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, then area of(△BPC) equal to
area of(△BPC) = area of (△DPQ)
It is given that ABCD is a parallelogram
Join point A to point C.
Consider △APC and △BPC
△APC and △BPC are lying on the same base PC and between the same parallels PC and AB. Therefore,
Area (△APC) = Area (△BPC) ... (1)
In quadrilateral ACQD, it is given that
AD = CQ
Since ABCD is a parallelogram,
CQ is a line segment which is obtained when line segment BC is produced.
AD || CQ
We have,
AC = DQ and AC ∥ DQ
Hence, ACQD is a parallelogram.
Consider △DCQ and △ACQ
These are on the same base CQ and between the same parallels CQ and AD. Therefore,
Area (△DCQ) = Area (△ACQ)
Area (△DCQ) - Area (△PQC) = Area (△ACQ) - Area (△PQC)
Area (△DPQ) = Area (△APC) ... (2)
From equations (1) and (2), we obtain
Area (△BPC) = Area (△DPQ)