Question

The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB = AC, is parallel to BC.

Options

• True

• False

Answer 1

This statement is True.

Explanation:

Let EAF be tangent to the circumcircle of ∆ABC.

To prove: EAF ॥ BC

We have, ∠EAB = ∠ACB   ...(i) [Angle between tangent and chord is equal to angle made by chord in the alternate segment]

Here, AB = AC

⇒ ∠ABC = ∠ACB  ...(ii)

From equation (i) and (ii), we get

∠EAB = ∠ABC

∵ Alternate angles are equal.

⇒ EAF ॥ BC