Question

In the adjoining figure, $XY$ is parallel to $AC$. If $XY$ divides the triangle into equal parts, then the value $\frac{AX}{AB}=$

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{\sqrt{2}+1}{\sqrt{2}}$
• D. $\frac{\sqrt{2}-1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{\sqrt{2}-1}{\sqrt{2}}$

We have, Area of $△BXY$ = Area of AXYC
Now adding Area of $△BXY$ on both sides, we get,
$2\times$ Area of $△BXY$=Area of $△ABC$
$\Rightarrow \frac{Areaof△ABC}{Areaof△BXY}=2$ ------(i)

In $△ABC$ and $△XBY$
$\angle ABC=\angle XBY$
$\angle BAC=\angle BXY$ ---------(corresponding angles)
By AA Similarity, $△ABC\sim △XBY$
From (i), we have,
$\frac{{AB}^2}{{BX}^2}=2$

$\Rightarrow \frac{AB}{BX}=\sqrt{2}$

$\Rightarrow \frac{BX}{AB}=\frac{1}{\sqrt{2}}$

$\Rightarrow 1-\frac{BX}{AB}=1-\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{AB-BX}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}$
$\Rightarrow \frac{AX}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}$