Question

In an isosceles triangle ABC with $AB=AC$ a circle passing through B and C intersect side AB and AC at D and E respectively. Prove that DE$\parallel$ BC.

Answer 1

In $△ABC$

$\angle B=\angle C.....\left(1\right)$

In the cyclic quadrilateral CBDE, side BD is produced to A.

We know that exterior angle is equal to opposite interior angle.

i.e $\angle D=\angle C....\left(2\right)$

From (1) and (2)

$\angle ADE=\angle ABC$

So, corresponding angles are equal

Hence, $DE||BC$