Question

D and E are points on the sides AB and AC respectively of a $△$ABC such that DE || BC and DE divides $△$ABC into two parts, equal in area. Find $\frac{BD}{AB}$.

• A. (√2−1)/√2
• B. (√2−2)/√2
• C. (√3−1)/2
• D. None of given options

Answer 1

The correct option is A

(√2−1)/√2

We have,

Area ($△$ADE) = Area (trapezium BCED)

Area of $△$ABC= Area of $△$ADE + Area of trapezium BCED

= Area of $△$ADE + Area of $△$ADE

= 2 Area of $△$ADE

In $△$ADE and $△$ABC, we have

$\angle$ ADE = $\angle$ B

[Since DE || BC $\angle$ ADE = $\angle$ B (Corresponding angles)]

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

So $△$ ADE ~ $△$ ABC

$\frac{areaof△ADE}{areaof△ABC}$ = $\frac{A{D}^2}{A{B}^2}$

$\frac{areaof△ADE}{2areaof△ADE}$ = $\frac{A{D}^2}{A{B}^2}$

Therefore $\frac{A{D}^2}{A{B}^2}$ = $\frac{1}{2}$

$\frac{AD}{AB}$ = $\frac{1}{\sqrt{2}}$

$\frac{BD}{AB}$ = $\frac{AB-AD}{AB}$

= 1 - $\frac{AD}{AB}$

= 1 - $\frac{1}{\sqrt{2}}$

= $\frac{\sqrt{2}-1}{\sqrt{2}}$ = $\frac{2-\sqrt{2}}{2}$