# 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 $\mathrm{AC}$. $]$"

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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.

Updated on 10-Oct-2022 13:46:28