Question : In triangle ABC, DE $\parallel$ BC where D is a point on AB and E is a point on AC. DE divides the area of $\Delta ABC$ into two equal parts. Then BD : AB is equal to:
Option 1: $\sqrt{2}:(\sqrt{2}+1)$
Option 2: $\sqrt{2}:(\sqrt{2}-1)$
Option 3: $(\sqrt{2}-1): \sqrt{2}$
Option 4: $(\sqrt{2}+1): \sqrt{2}$
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Correct Answer: $(\sqrt{2}-1): \sqrt{2}$
Solution :
Given: DE $\parallel$ BC
According to the question,
$area(\triangle ADE)=area(\operatorname{trapezium BCED})$
⇒ $area(\triangle ADE)+area(\triangle ADE)=area(\operatorname{trapezium BCED})+area(\triangle ADE)$
⇒$2area(\triangle ADE)=area(\triangle ABC)$.............(i)
From the figure,
$BD=AB-AD$
⇒ $AD=AB-BD$..................(ii)
In $\triangle ADE$ and $\triangle ABC$
$\angle D=\angle D$ (corresponding angle)
$\angle A=\angle A$ (common angle)
So, $\triangle ADE\sim\triangle ABC$
⇒ $\frac{area(\triangle ADE)}{area(\triangle ABC)}=\frac{AD^2}{AB^2}$
From equation (i)
$\frac{area(\triangle ADE)}{2area(\triangle ADE)}=\frac{AD^2}{AB^2}$
⇒ $\frac{AD^2}{AB^2}=\frac{1}{2}$
⇒ $\frac{AD}{AB}=\frac{1}{\sqrt2}$
⇒ $AB=\sqrt2AD$
From equation (ii)
$AB=\sqrt2(AB-BD)$
⇒ $\sqrt2BD=\sqrt2AB-AB$
⇒ $\sqrt2BD=AB(\sqrt2-1)$
⇒ $\frac{BD}{AB}=\frac{\sqrt2-1}{\sqrt2}$
Hence, the correct answer is $(\sqrt2-1):\sqrt2$.
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