Question

In figure, the line segment XY is parallel to side AC of $\Delta ABC$ and it divides the triangle into two parts of equal areas. Find the ratio $\frac{AX}{AB}$.

Answer 1

REF.Image

2 Area (xy) = Area (ABC)

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

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

$\angle ACB=\angle XYB$ (since XY||AC, angles are equal)

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

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

$(AB{)}^2=2(XB{)}^2$ $AB=\sqrt{2}\times B$

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

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