Question

$ABCD$ is a parellelogram, $AB$ is divided at $P$ and $CD$ divided at $Q$. So that $AP:PB=3:2$ and $CQ:QD=4:1$. If $PQ$ intersect $AC$ at$R$, then prove that $AR=3/7AC$.

Answer 1

Since $ABCD$ is a parallelogram then $AB\parallel CD$ and $AD\parallel BC$

Since $AB\parallel CD$ and $AC$ is a transversal, so

$\angle PAR=\angle QCR$(alternate interior angles are equal).....$\left(1\right)$

Since $ABCD$ is a parallelogram,then $AB=CD$ and $AD=BC$, as opposite sides of parallelogram are equal.

let $AB=CD=x$

So, $AP=\frac{3x}{5}$ and $PB=\frac{2x}{5}$ as $AP:PB=3:2$

and $CQ=\frac{4x}{5}$ and $QD=\frac{x}{5}$ as $CQ:QD=4:1$

In $△CQR$ and $△PAR$

$\angle QCR=\angle PAR$ using $\left(1\right)$

and $\angle QRCR=\angle PRA$ (vertically opposite angles)

$\Rightarrow △CQR\sim △PAR$ by AA similarity.

$\Rightarrow \frac{CR}{AR}=\frac{CQ}{AQ}=\frac{QR}{PR}$(corresponding sides of similar triangles are proportional)

$\Rightarrow \frac{CR}{AR}=\frac{CQ}{AQ}=\frac{\frac{4x}{5}}{\frac{3x}{5}}=\frac{4}{3}$

$\Rightarrow \frac{CR}{AR}+1=\frac{4}{3}+1=\frac{7}{3}$

$\Rightarrow \frac{CR+AR}{AR}=\frac{CA}{AR}=\frac{7}{3}$

$\Rightarrow 7AR=3AC$

$\Rightarrow AR=\frac{3}{7}AC$