Question

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

Answer 1

We are given that XY || AC.

$\Rightarrow$ $\angle$1 = $\angle$3 and $\angle$2 = $\angle$4 [ corresponding angles ]

$\Rightarrow$ $△$BXY ~ $△$BAC [ AA similarity ]

$\Rightarrow$ $\frac{ar\left(△BXY\right)}{ar\left(△BAC\right)}=\frac{(BX{)}^2}{(BA{)}^2}$ [ By theorem] ..... (i)

Also, we are given that

ar ($△$ BXY ) = ar ($△$ BAC ) $\Rightarrow$ $\frac{ar\left(△BXY\right)}{ar\left(△BAC\right)}={\left(\frac{1}{2}\right)}^{\frac{1}{2}}$ ....(ii)