Question

$\text{In the following figure, triangle AXY}\text{is isosceles with}\angle AXY=\angle AYX.$

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

Answer 1

$\text{Here,}\frac{BX}{AX}=\frac{CY}{AY}$

$\Rightarrow \frac{AX}{BX}=\frac{AY}{CY}$
$\text{(By taking reciprocals on both sides)}$

$\therefore XY||BC$
$\text{(By converse of basic proportionality}\text{theorem)}$

$\therefore \angle B=\angle AXY\text{and}\angle C=\angle AYX$
$\text{(Corresponding angles)}$

$\text{But}\angle AXY=\angle AYX\text{(Given)}$

$\Rightarrow \angle B=\angle C$

$\Rightarrow AC=AB\text{(Sides opposite to equal angles are}\text{equal)}$

$\therefore \Delta ABC\text{is an isosceles triangle.}$