Question

In the given figure, $△ABC$ is an isosceles triangle in which $AB=AC$ and a circle passing through $B$ and $C$ intersect $AB$ and $AC$ at $D$ and $E$ respectively. Prove that $DE||BC$.

Answer 1

We know that $△ABC$ is an isosceles triangle in which $AB=AC$ and a circle passing through $B$ and $C$ intersects $AB$ and $AC$ at $D$ and $E$

So $AB=AC$

We get

$\angle ACB=\angle ABC$

It can be written as

$\angle ADE=\angle ACB=\angle ABC$

So we get

$\angle ADE=\angle ABC$

$DE\parallel BC$

Therefore, it is proved that $DE\parallel BC$.