Question : In $\triangle$ABC, D and E are points on AB and AC, respectively, such that DE || BC and DE divide the $\triangle$ABC into two parts of equal areas. The ratio of AD and BD is:
Option 1: $1:1$
Option 2: $1:(\sqrt2-$1)
Option 3: $1:\sqrt2$
Option 4: $1:(\sqrt2+$1)
Correct Answer: $1:(\sqrt2-$1)
Solution :
Area($\triangle$ADE) = Area(trapezium BCED)
⇒ Area($\triangle$ADE) + Area($\triangle$ADE) = Area(trapezium BCED) + Area($\triangle$ADE)
⇒ 2 × Area($\triangle$ADE) = Area($\triangle$ABC)
In $\triangle$ADE and $\triangle$ABC,
$\angle$ADE = $\angle$ABC [i.e., corresponding angles in DE || BC]
$\angle$A = $\angle$A [common angle]
⇒ $\triangle$ADE ~ $\triangle$ABC
⇒ $\frac{\text{Area($\triangle$ADE)}}{\text{Area($\triangle$ABC)}}$ = $\frac{AD^2}{AB^2}$
⇒ $\frac{\text{Area($\triangle$ADE)}}{\text{2×Area($\triangle$ADE)}}$ = $\frac{AD^2}{AB^2}$
⇒ $\frac{1}{2}$ = $\frac{AD^2}{AB^2}$
⇒ $\frac{AD}{AB}$ = $\frac{1}{\sqrt2}$------------------(i)
⇒ AB = $\sqrt2$AD
⇒ AD + BD = $\sqrt2$AD
⇒ BD = $(\sqrt2-1)$AD
⇒ AD : BD = $1:(\sqrt2-1)$
Hence, the correct answer is $1:(\sqrt2-1)$.
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