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).

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Solution

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.

$Δ A P C a n d Δ B P C$ are lying on the same base PC and between the same parallels PC and AB.
Therefore, $a r (Δ A P C) = a r (Δ B P C) . . . (1)$

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.
$∴ A D | | C Q$
We have,
AC = DQ and AC|| DQ
Hence, ACQD is a parallelogram.

Consider
$Δ D C Q a n d Δ A C Q$
These are on the same base CQ and between the same parallels CQ and AD. Therefore,
$a r (Δ D C Q) = a r (Δ A C Q)$

Then, $a r (Δ D C Q) - a r (Δ P Q C) = a r (Δ A C Q) - a r (Δ P Q C)$ [Subtracting $A r e a (Δ P Q C)$ from both sides]
$Δ a r (Δ D P Q) = a r (Δ A P C) . . . (2)$

From equations (1) and(2), we obtain
$a r (Δ B P C) = a r (Δ D P Q)$

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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).

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).