Question

In triangle $ABC,XY||AC$ and divide the triangle into two parts of equal areas. Find the ratio $\frac{AX}{AB}$.

Answer 1

In $\Delta ABC$

$XY$ is parallel to $AC$

To find : $\frac{AX}{AB}$

In $\Delta ABC$ & $\Delta XBY$

$\angle ABC=\angle XBY$

$\angle ACB=\angle XYB$

$\Delta ABC\sim \Delta XBY$ [AA similarity]

Now,

We know that in similar triangles,

Ratio of area of triangle is equal to ratio of square of corresponding sides

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

$\frac{\text{Area of}\Delta XBY+\text{Area of}XYCA}{\text{Area of}\Delta XBY}={\left(\frac{AB}{XB}\right)}^2$

It is given that $XY$ divides triangle in two equal parts.

$\frac{\text{Area of}\Delta XBY+\text{Area of}\Delta XBY}{\text{Area of}\Delta XBY}={\left(\frac{AB}{XB}\right)}^2$

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

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

Now, we need to find $\frac{AX}{AB}$

Since $AX+XB=AB$

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

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

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

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