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 that are equal in area. Find $\frac{BD}{AB}$.

• A. $\frac{\sqrt{2}-1}{\sqrt{2}}$
• B. $\frac{2-\sqrt{2}}{2}$
• C. (a) and (b) above.
• D. None of the above

Answer 1

The correct option is C (a) and (b) above.

We have,

Area ($△$ADE) = Area (trapezium BCED)

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

= Area of $△$ADE + Area of $△$ADE

= 2$\times$ Area of $△$ADE

In $△$ADE and $△$ABC, we have

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

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

So, $△$ ADE ~ $△$ ABC [by AA similarity criterion]

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

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

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

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

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

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

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

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

Hence, both options (A) and (B) are correct.