Question

In the given figure, the line segment $XY$ is parallel in $AC$ of $\Delta ABC$ and is divided the triangle into two parts of equal areas then $\frac{AX}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}$

• A. True
• B. False

Answer 1

The correct option is A True

In $△ABC,XY\parallel AC$

Area of $△ABC=2\times Areaof△BXY$ ........$\left(1\right)$

$XY\parallel AC$ and $BA$ is transversal.

$\therefore \angle BXY=\angle BAC$ .......$\left(2\right)$

$\angle XBY=\angle ABC$(Common angle)

$△BXY\cong △BXY$

$\Rightarrow \frac{Areaof△BAC}{Areaof△BXY}=2$

$\Rightarrow \frac{{BA}^2}{{BX}^2}=2$

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

$\Rightarrow BA=\sqrt{2}BX$

$\Rightarrow BA=\sqrt{2}(AB-AX)$

$\Rightarrow (1-\sqrt{2})BA=-\sqrt{2}AX$

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

Hence the given statement is true.