In the figure, $ABCD$ is a parallelogram in which $P$ is the mid-point of $DC$ and $Q$ is a point on $AC$ such that $CQ = \frac{1}{4}AC$. If $PQ$ produced meets $BC$ at $R$, prove that $R$ is a mid-point of $BC$.
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Given:

$ABCD$ is a parallelogram in which $P$ is the mid-point of $DC$ and $Q$ is a point on $AC$ such that $CQ = \frac{1}{4}AC$.

$PQ$ produced meets $BC$ at $R$.

To do:

We have to prove that $R$ is a mid-point of $BC$.

Solution:

Join $BD$.

Diagonals $AC$ and $BD$ bisect each other at $O$.

This implies,

$AO = OC = \frac{1}{2}AC$.....…(i)

In $\triangle OCD$,

$P$ and $Q$ are the mid-points of $CD$ and $CO$.

This implies,

$PQ \parallel OD$ and $PQ = \frac{1}{2}OD$

In $\triangle BCD$,

$P$ is the mid-point of $DC$ and $PQ \parallel OD$.

This implies,

$PR \parallel BD$

Therefore,

$R$ is the mid-point of $BC$.

Hence proved.

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Updated on: 10-Oct-2022

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