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

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Solution

We have,

$A r e a (△ A D E) = A r e a (t r a p e z i u m B C E D)$

$\Rightarrow$ $A r e a (△ A D E) + A r e a (△ A D E) = A r e a (t r a p e z i u m B C E D) + A r e a (△ A D E)$

$\Rightarrow$ $2 A r e a (△ A D E) = A r e a (△ A B C) . . . . . . (i)$

In $△ A D E$ and $△ A B C$ , we have

$∠ A D E = ∠ B$ [ $∵ D E | | B C ∴ ∠ A D E = ∠ B$ (Corresponding angles]

and, $∠ A = ∠ A$ [Common]

$∴$ $△ A D E \sim △ A B C$

$\Rightarrow$ $\frac{A r e a (△ A D E)}{A r e a (△ A B C)} = \frac{{A D}^{2}}{{A B}^{2}}$

$\Rightarrow$ $\frac{A r e a (△ A D E)}{2 A r e a (△ A B C)} = \frac{{A D}^{2}}{{A B}^{2}}$

$\Rightarrow$ $\frac{1}{2} = {(\frac{A D}{A B})}^{2}$

$\Rightarrow$ $\frac{A D}{A B} = \frac{1}{\sqrt{2}}$

$\Rightarrow$ $A B = \sqrt{2} A D$

$\Rightarrow$ $A B = \sqrt{2} (A B - B D)$

$\Rightarrow$ $(\sqrt{2} - 1) A B = \sqrt{2} B D$

$\Rightarrow$ $\frac{B D}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}} = \frac{2 - \sqrt{2}}{2}$

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