In an isosceles triangle ABC with AB = AC a circle passing through B and C intersect the sides AB and AC at D and E respectively.Prove that DE||BC.

**To prove:** DE || BC.

**Proof:**

In order to prove that DE || BC it is sufficient to show that ∠B = ∠ADE

In ΔABC,

AB = AC

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

It is the property of cyclic quadrilateral that its exterior angle is equal to the opposite interior angle.

⇒ ∠ADE = ∠C .... (2)

From (1) and (2), we get

, which forms corresponding angles.

Hence DE || BC.

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