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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