Answer to Question #94870 in Geometry for Clifford Sekyi Bosompem

Question

In a parallelogram ABCD, P divides AB in the ratio 2:5 and Q divides DC in the ratio 3:2. If AC and PQ intersect at R , find the ratios AR:RC and PR:RQ

Answer 1

1)$AB||CD$, $PQ$ is transversal line, $\angle RPA$ and $\angle CQR$ are alternate interior angles, so $\angle RPA=\angle CQR$

2)$AB||CD$, $AC$ is transversal line, $\angle PAR$ and $\angle QCR$ are alternate interior angles, so $\angle PAR=\angle QCR$

3)$\angle PRA=\pi-\angle RPA-\angle PAR$ in $\Delta APR$ , $\angle CRQ=\pi-\angle CQR-\angle CRQ$ in $\Delta CRQ$

$\angle PRA=\pi-\angle RPA-\angle PAR=\pi-\angle CQR-\angle CRQ=\angle CRQ$

So $\Delta APR \sim \Delta CQR$ and we have $\frac{AP}{CQ}=\frac{AR}{CR}=\frac{PR}{QR}$

Since $\frac{AP}{BP}=\frac{2}{5}$, we obtain that $AP=2x$ and $BP=5x$ for some $x$

Similarly we obtain that $DQ=3y$ and $QC=2y$ for some $y$

So $7x=AP+BP=AB=CD=CQ+QD=5y$ , and $y=1.4x$

We have $\frac{AR}{CR}=\frac{PR}{QR}=\frac{AP}{CQ}=\frac{2x}{2y}=\frac{2x}{2\cdot 1.4x}=\frac{5}{7}$

Answer: $\frac{AR}{CR}=\frac{PR}{QR}=\frac{5}{7}$