Question

In given figure, If $\frac{AD}{DC}=\frac{BE}{EC}$ and $\angle CDE=\angle CED$, prove that $△CAB$ is isosceles.

Answer 1

In $△ABC$, we have

$\frac{AD}{DC}=\frac{BE}{EC}$ [Given]

Therefore, by the converse of basic proportionality theorem, we have,

$DE||AB$

$\Rightarrow$ $\angle CDE=\angle CAB$and $\angle CED=\angle CBA$ [Corresponding angles]

But, $\angle CDE=\angle CED$ [Given]

$\therefore$ $\angle CAB=\angle CBA$

$\Rightarrow$ $\angle A=\angle B$

$\Rightarrow$ $BC=AC$ [$\because$ Sides opposite to equal angles are equal]

$\Rightarrow$ $△CAB$ is isosceles.