Question

In the given fig, the line segment $xy$ in parallel to $AC$ of $△ABC$ and its dividend the triangle into two points of equal areas. Prove that $\frac{Ax}{AB}=\frac{\sqrt{2-1}}{\sqrt{2}}$

Answer 1

Given in $\Delta ABC$, $xy$ is parallel to $AC$ and $areaof\Delta XBY=areaofAXYC$

In $\Delta ABC$ and $\Delta XBY$,

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

$\angle ACB=\angle XYB$ (corresponding angles)

$\Delta ABC\sim \Delta XBY$ (AA similarity)

We know that in similar triangles, ratio of area of triangle is equal to ratio of square of corresponding sides.

$\frac{Areaof\Delta ABC}{Areaof\Delta XBY}={\left(\frac{AB}{XB}\right)}^2$

$\Rightarrow \frac{Areaof\Delta XBY+AreaofAXYC}{Areaof\Delta XBY}={\left(\frac{AB}{XB}\right)}^2$

$\Rightarrow \frac{Areaof\Delta XBY+Areaof\Delta XBY}{Areaof\Delta XBY}={\left(\frac{AB}{XB}\right)}^2$ (given)

$\Rightarrow \frac{2\times Areaof\Delta XBY}{Areaof\Delta XBY}={\left(\frac{AB}{XB}\right)}^2$

$\Rightarrow 2={\left(\frac{AB}{XB}\right)}^2$

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

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

Now let us find $\frac{AX}{AB}$

Since $AX+XB=AB$

$\Rightarrow AX+\frac{AB}{\sqrt{2}}=AB$

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

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

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