In the adjoining figure, ABC is an isosceles triangle in which AB = AC and circle passing through B and C intersects sides AB and AC at points D and E. Prove that DE || BC.

Given, AB = AC

∴ ∠ABC = ∠ACB (As angles opposite to equal sides are equal)

As BCED is a cyclic quadrilateral,

∠ADE = ∠BCE (∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.)

∴ ∠ADE = ∠ABC

Since these angles are equal and are corresponding this is the property of parallel lines,

∴ DE || BC.

Hence, proved that DE || BC.