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# In figure below, $ \mathrm{ABCD} $ is a parallelogram and $ \mathrm{BC} $ is produced to a point $ \mathrm{Q} $ such that $ \mathrm{AD}=\mathrm{CQ} $. If $ \mathrm{AQ} $ intersect $ \mathrm{DC} $ at $ \mathrm{P} $, show that ar $ (\mathrm{BPC})= $ ar $ (\mathrm{DPQ}) $.

[Hint : Join $ \mathrm{AC} $. $ ] $

"

Given:

$D$ and \( \mathrm{E} \) are two points on \( \mathrm{BC} \) such that \( \mathrm{BD}=\mathrm{DE}=\mathrm{EC} \).

To do:

We have to show that \( \operatorname{ar}(\mathrm{ABD})=\operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{AEC}) \).

Solution:

In $\triangle ADP$ and $\triangle QCP$,

$\angle APD = \angle QPC$ (Vertically opposite angles)

$\angle ADP = \angle QCP$ (Alternate angles)

$AD = CQ$

Therefore, by AAS congruency,

$\triangle ADP \cong \triangle QCP$

This implies,

$DP = CP$ (CPCT)

In $CDQ$,

$DP=CP$

This implies,

$QP$ is the median.

We know that,

The median of a triangle divides it into two parts of equal areas.

This implies,

$ar(\triangle DPQ) = ar(\triangle QPC)$...........(i)

$AD = CQ$ ($\triangle ADP \cong \triangle QCP$)

$AD = BC$ ($ABCD$ is a parallelogram)

This implies,

$BC = QC$

In $\triangle PBQ$,

$BC=QC$

$PC$ is the median.

This implies,

$ar(\triangle QPC) = ar(\triangle BPC)$............(ii)

From (i) and (ii), we get,

$ar(\triangle BPC) = ar(\triangle DPQ)$

Hence proved.

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