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$\mathrm{ABCD}$ is a parallelogram, $\mathrm{P}$ is the mid-point of $\mathrm{AB}$. If $\mathrm{R}$ is the point of intersection
of $\mathrm{AC}$ and $\mathrm{DP}$, then $\mathrm{R}$ divides $\mathrm{AC}$ internally in the ratio
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of $\mathrm{AC}$ and $\mathrm{DP}$, then $\mathrm{R}$ divides $\mathrm{AC}$ internally in the ratio
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Verified Answer
The correct answer is:
$1: 2$
Draw BS parallel to DP as shown
Let $A P=P B=x \Rightarrow D S=x \Rightarrow S C=x$
$\triangle B A Q \sim \triangle P A R$
$\therefore \frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{AQ}}{\mathrm{AR}} \Rightarrow \frac{2 \mathrm{x}}{\mathrm{x}}=\frac{\mathrm{AQ}}{\mathrm{AR}} \Rightarrow \mathrm{AQ}=2 \mathrm{AR}$
Thus $\mathrm{R}$ is mid point of $\mathrm{AQ}$. i.e. $\mathrm{AR}=\mathrm{RQ}$
Similarly $\Delta$ CQS $\Delta$ CRD
$\therefore \mathrm{CQ}=\mathrm{RQ}$
Thus we get $\mathrm{AR}=\mathrm{RQ}=\mathrm{CQ}$
Hence point $R$ divides $A C$ in the ratio $1: 2$
Let $A P=P B=x \Rightarrow D S=x \Rightarrow S C=x$
$\triangle B A Q \sim \triangle P A R$
$\therefore \frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{AQ}}{\mathrm{AR}} \Rightarrow \frac{2 \mathrm{x}}{\mathrm{x}}=\frac{\mathrm{AQ}}{\mathrm{AR}} \Rightarrow \mathrm{AQ}=2 \mathrm{AR}$
Thus $\mathrm{R}$ is mid point of $\mathrm{AQ}$. i.e. $\mathrm{AR}=\mathrm{RQ}$
Similarly $\Delta$ CQS $\Delta$ CRD
$\therefore \mathrm{CQ}=\mathrm{RQ}$
Thus we get $\mathrm{AR}=\mathrm{RQ}=\mathrm{CQ}$
Hence point $R$ divides $A C$ in the ratio $1: 2$
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