Question

In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD =CQ. If AQ intersects DC at P,
Show that ar (BPC)= ar (DPQ).

Answer 1

It is given that ABCD is a parallelogram.
AD || BC and AB || DC (Opposite sides of a parallelogram are parallel to each other)
Join point A to point C.

$\Delta APCand\Delta BPC$ are lying on the same base PC and between the same parallels PC and AB.
Therefore, $ar\left(\Delta APC\right)=ar\left(\Delta BPC\right)...\left(1\right)$

In quadrilateral ACQD, AD = CQ
AD || BC (Opposite sides of a parallelogram are parallel)
CQ is a line segment which is obtained when line segment BC is produced.
$\therefore AD||CQ$
We have,
AC = DQ and AC|| DQ
Hence, ACQD is a parallelogram.

Consider
$\Delta DCQand\Delta ACQ$
These are on the same base CQ and between the same parallels CQ and AD. Therefore,
$ar\left(\Delta DCQ\right)=ar\left(\Delta ACQ\right)$

Then, $ar\left(\Delta DCQ\right)-ar\left(\Delta PQC\right)=ar\left(\Delta ACQ\right)-ar\left(\Delta PQC\right)$ [Subtracting $Area\left(\Delta PQC\right)$ from both sides]
$\Delta ar\left(\Delta DPQ\right)=ar\left(\Delta APC\right)...\left(2\right)$

From equations (1) and(2), we obtain
$ar\left(\Delta BPC\right)=ar\left(\Delta DPQ\right)$