Given: P is a point outside the circle with centre O. From point P, two tangents PA and PB are drawn, whose points of contact are A and B respectively. O, A; O, B; and O, P are joined. Thus, PA and PB subtend ∠POA and ∠POB at the centre respectively.

To prove that: (i) PA = PB (ii) ∠POA = ∠POB

Proof: PA and PB are tangents, and OA and OB are the radii through the points of contact.

∴ OA ⊥ PA and OB ⊥ PB

In the right-angled triangles POA and POB,

∠OAP = ∠OBP (each is 90°)

The hypotenuse OP is the common side, and OA = OB (radii of the same circle).

∴ ΔPOA ≅ ΔPOB (by R-H-S axiom of congruency)

Hence,

PA = PB (corresponding sides) …… (i) proved

∠POA = ∠POB (corresponding angles) …… (ii) proved

From this theorem we get:

∠APO = ∠BPO (corresponding angles)

∴ OP bisects ∠APB.