Question

D and E are points on the sides AB and AC respectively of a $△$ABC such that DE || BC

and divides $△$ABC into two parts, equal in area. Find $\frac{BD}{AB}$

Answer 1

We have, Area ($△$ADE) = Area (trapezium BCED) $\Rightarrow$ Area ($△$ADE) + Area ($△$ADE)

= Area (trapezium BCED) + Area ($△$ADE)

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

In $△$ADE and $△$ABC, we have

$\angle$ADE = $\angle$B

[i.e, DE II BC $\therefore$ $\angle$ADE = $\angle$B (Corresponding angles)]

and, $\angle$A = $\angle$A [Common]

$\therefore$ $\angle$ADE ~ $\angle$ABC

$\Rightarrow$ $\frac{Areaof\left(△ADE\right)}{Areaof\left(△ABC\right)}=\frac{A{D}^2}{A{B}^2}$

$\Rightarrow$ $\frac{Areaof\left(△ADE\right)}{2Areaof\left(△ADE\right)}=\frac{A{D}^2}{A{B}^2}$

$\Rightarrow$ $\frac{1}{2}={\left(\frac{AD}{AB}\right)}^2$ $\Rightarrow$ $\frac{AD}{AB}=\frac{1}{\sqrt{2}}$

$\Rightarrow$ AB = $\sqrt{2}$ AD

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

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

$\Rightarrow$ $\frac{BD}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{2-\sqrt{2}}{2}$