Question

ABCD is a parallelogram . AB is divided at P and CD at Q so that AP : PB = 3 : 2 and CQ : QD = 4 : 1 . If PQ meets AC at R then prove that AR = $\frac{3}{7}$AC .

Answer 1

Let, AB$=$CD$=5x$ according to the parallelogram definition that opposite sides are equal.

$AP=3x,PB=2x$ and $CQ=4x$, $QD=x$

$\Rightarrow ARP\sim CRQ$

$\frac{3x}{4x}=\frac{AR}{RC}$

$\frac{AR}{RC}=\frac{3}{4}$

According to this fraction, we have

$AR=3$ & $RC=4$ so that $AC=3+4=7$

Finally, $\frac{AR}{AC}=\frac{3}{7}$

$\therefore AR=\frac{3}{7}AC$.