Question

In the following figure, $\angle AXY=\angle AYX$.

If $\frac{BX}{AX}=\frac{CY}{AY}$, Prove that triangle ABC is an isosceles.

Answer 1

In the given figure,

$\angle AXY=\angle AYX$

and $\frac{BX}{AX}=\frac{CY}{AY}$

In $\Delta AXY$

$\angle AXY=\angle AYX$ (given)

$\therefore AY=AX$ (Sides opposite to equal angles)

$\frac{BX}{AX}=\frac{CY}{AY}$

$\Rightarrow \frac{AX}{BX}=\frac{AY}{CY}$

$\therefore XY||BC$

$\therefore \angle B=\angle AXY$ and $\angle C=\angle AYX$

(Corresponding angles)

But $\angle AXY=\angle AYX$ is given

$\therefore \angle B=\angle C$

$\therefore AC=AB$ (Side opposite to equal angles)

$\therefore \Delta ABC$ is an isosceles triangle.