ABCD is a parallelogram,P is the mid-point of | vedclass

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(C) In $	riangle ABP$ and $	riangle CDP$,$AB \parallel DC$ and $AB = DC$. Since $P$ is the mid-point of $AB$,$AP = PB = \frac{1}{2} AB = \frac{1}{2}

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$1: 2$

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(C) In $	riangle ABP$ and $	riangle CDP$,$AB \parallel DC$ and $AB = DC$. Since $P$ is the mid-point of $AB$,$AP = PB = \frac{1}{2} AB = \frac{1}{2} DC$.Consider $	riangle APR$ and $	riangle CPD$:$\angle PAR = \angle PCD$ (alternate interior angles as $AB \parallel DC$)$\angle APR = \angle CPD$ (vertically opposite angles)Thus,$	riangle APR \sim 	riangle CPD$ by $AA$ similarity.Therefore,the ratio of corresponding sides is equal:$\frac{AR}{CR} = \frac{AP}{CD} = \frac{\frac{1}{2} AB}{AB} = \frac{1}{2}$.Hence,$R$ divides $AC$ in the ratio $1: 2$.

$ABCD$ is a parallelogram,$P$ is the mid-point of $AB$. If $R$ is the point of intersection of $AC$ and $DP$,then $R$ divides $AC$ internally in the ratio

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