If two tangents are drawn to a circle from a point outside it, then prove that the line segments joining the point of contact and the exterior point are equal and they subtend equal angles at the centre.
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.
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