Question

PAQ is a tangent at A to the circumcircle of ΔABC such that PAQ is parallel to BC, prove that ΔABC is an isosceles triangle.

Answer 1

Given:

PAQ is a tangent to the circumcircle of ΔABC at A.

PAQ ∥ BC.

Since PAQ is tangent at A to the circle, by the Alternate Segment Theorem:

∠PA B = ∠ACB

∠CAQ = ∠ABC

But PAQ ∥ BC, so:

∠PAB = ∠ABC (corresponding angles)

∠CAQ = ∠ACB (corresponding angles)

So from the tangent theorem:

∠ACB = ∠PAB = ∠ABC

Thus,

∠ACB = ∠ABC

Therefore, in ΔABC, the base angles at B and C are equal.

Hence, AB = AC