Question

P and Q are points on the sides AB and AC respectively of $△$ABC such that PQ||BC and divides $△$ABC into two parts, equal in area. Find PB : AB.

Answer 1

Area ($△$APQ) = Area (trap. PBCQ) [Given]

$\Rightarrow$ Area ($△$APQ) = [Area ($△$ABC) - Area ($△$APQ)]

$\Rightarrow$ 2 Area ($△$APQ) = ($△$ABC)

$\Rightarrow$ $\frac{Areaof\left(△APQ\right)}{Areaof\left(△ABC\right)}$ = $\frac{1}{2}$ ......(i)

Now, in $△$APQ and $△$ABC , we have

$\angle$PAQ = $\angle$BAC [Common $\angle$A ]

$\angle$APQ = $\angle$ABC [PQBC, corresponding $\angle$s are equal ]

$\therefore$ $△$APQ ~ $△$ABC.

We know that the areas of similar $△$s are proportional to the squares of their corresponding sides

$\therefore$ $\frac{Areaof\left(△APQ\right)}{Areaof\left(△ABC\right)}=\frac{A{P}^2}{A{B}^2}$ $\Rightarrow$ $\frac{A{P}^2}{A{B}^2}=\frac{1}{2}$ [ Using (i) ]

$\Rightarrow$ $\frac{AP}{AB}=\frac{1}{\sqrt{2}}$ i.e., AB = $\sqrt{2}$ AP

$\Rightarrow$ AB = $\sqrt{2}$ (AB - PB) $\Rightarrow$ $\sqrt{2}$ PB = ($\sqrt{2}$ - 1) AB

$\Rightarrow$ $\frac{PB}{AB}$ =$\frac{(\sqrt{2}-1)}{\sqrt{2}}$