Question

In $△ABC$, $AP:PB=2:3$. $PO\parallel BC$ and is extended to $Q$ so that $CQ\parallel BA$. Find $area\left(△APO\right):area\left(△ABC\right)$

Answer 1

In $△ABC$,$PO\parallel BC$.

using Basic Proportionality theorem,

$\frac{AP}{PB}=\frac{AO}{OC}$

$\Rightarrow \frac{AO}{OC}=\frac{2}{3}$ ........$\left(1\right)$

$\angle PAO=\angle BAC$(common)

$\angle PAO=\angle ABC$(corresponding angles)

$△APO\sim △ABC$ by AA similarity

$\therefore \frac{Ar\left(△APO\right)}{Ar\left(△ABC\right)}={\left(\frac{AO}{AC}\right)}^2={\left(\frac{2}{2+3}\right)}^2=\frac{4}{25}$