Question

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 show that area of($△$BPC) = area of ($△$DPQ).

Answer 1

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 $\parallel$ 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)