Question

In a $△ABC,D$ and $E$ are points on $AB$ and $AC$ respectively such that $\frac{AD}{DB}=\frac{AE}{EC}$ and $\angle ADE=\angle DEA$. Prove that $△ABC$ is isosceles

Answer 1

Since $\frac{AD}{DB}=\frac{AE}{EC}$, by converse of Thales theorem, $DE\parallel BC$
$\therefore \angle ADE=\angle ABCand...\left(1\right)$ and
$\angle DEA=\angle BCA....\left(2\right)$
But, given that $\angle ADE=\angle DEA...\left(3\right)$
From $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get $\angle ABC=\angle BCA$
$\therefore AC=AB$ (If opposite angles are equal, then opposite sides are equal).
Thus, $△ABC$ is isosceles.