Question : In a $\triangle ABC$, D and E are two points on AB and AC respectively such that DE || BC, DE bisects the $\triangle ABC$ in two equal areas. Then the ratio DB : AB is:
Option 1: $1:\sqrt2$
Option 2: $1:2$
Option 3: $\left ( \sqrt2-1 \right ):\sqrt2$
Option 4: $\sqrt2:1$
Correct Answer: $\left ( \sqrt2-1 \right ):\sqrt2$
Solution :
Given, area of ($\triangle ADE)$ = area of $BDEC$
⇒ $\triangle 2ADE = \triangle ABC$
Since, $DE \parallel BC$
∴ $\triangle ADE$ ~ $\triangle ABC$
Now,
$\frac{area (\triangle ADE)}{area (\triangle ABC)} = \frac{1}{2} = \frac{AD^2}{AB^2}$
⇒ $\frac{AB}{AD} = \sqrt2$....................(1)
⇒ $\frac{AB}{AD}-1 = \sqrt2-1$
⇒ $\frac{AB-AD}{AD}= \sqrt2-1$
⇒ $\frac{BD}{AD} = \sqrt2-1$......................(2)
$\therefore \frac{BD}{AB} = \frac{BD}{AD}× \frac{AD}{AB}= \frac{\sqrt2-1}{\sqrt2}$
Hence, the correct answer is $(\sqrt2-1):\sqrt2$.
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