Given:
P is an external point to a circle with centre O. PA and PB are the tangents from P to the circle. A and B are the points of contact.
To prove:
PA = PB
Construction:
Join OA, OB, OP.
Proof:
In triangle APO and BPO,
From the above theorem,
1. ∠AOP = ∠BOP (CPCT) This states that the two tangents subtend equal angles at the centre of the circle
2. ∠APO = ∠BPO (CPCT) The tangents are equally inclined to the line joining the point and the centre of the circle.
Or the centre of the circle lies on the angle bisector of the ∠APB.
