ABC is an isoscles triangle in which AB=AC. A circle passing through B and C intersects the sides AB and AC at D and E respectively then show that DE is parallel to BC?

Jun 1, 2018

see explanation.

Explanation:

Given $A B = A C , \implies \angle A B C = \angle A C B$
Let $\angle A B C = x , \implies \angle A C B = x$
As $B C E D$ is a cyclic quadrilateral, opposite angles add up to ${180}^{\circ}$,
$\implies \angle B D E = 180 - \angle E C B = 180 - x$,
$\implies A D E = 180 - \angle B D E = 180 - \left(180 - x\right) = x$
as $\angle A D E = \angle A B C = x$,
$\implies D E$ // $B C$