Question

In given figure, XY||AC and XY divides triangular region ABC into two parts equal in area. Determine $\frac{AX}{AB}$.

Answer 1

We have XY||AC

and, $Area\left(△BXY\right)=Area\left(quadXYCA\right)$

$\Rightarrow$ $Area\left(△ABC\right)=2Area\left(△BCY\right)$........(i)

Now, $XY||AC$ and BA is a transversal.

$\Rightarrow$ $\angle BXY=\angle BAC$.......(ii)

Thus, in ${△}^{\prime }sBAC$ and $BXY$, we have

$\angle XBY=\angle ABC$ [Common]

$\angle BXY=\angle BAC$ [From(ii)]

Therefore, AA-criterion of similarity, we have

$△BAC\sim △BXY$

$\Rightarrow$ $\frac{Area\left(△BAC\right)}{Area\left(△BXY\right)}=\frac{{BA}^2}{{BX}^2}$

$\Rightarrow$$2=\frac{{BA}^2}{{BX}^2}$ [Using (i)]

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

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

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