Question

TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.

Answer 1

Join PB.

In ΔTAP and ΔTBP,

TA = TB (tangents segments from an external points are equal in length)

Also, ∠ATP = ∠BTP. (since OT is equally inclined with TA and TB) TP = TP (common)

ΔTAP ≅ ΔTBP (by SAS criterion of congruency)

∠TAP = ∠TBP (corresponding parts of congruent triangles are equal)

But ∠TBP = ∠BAP (angles in alternate segments)

Therefore, ∠TAP = ∠BAP.

Hence, AP bisects ∠TAB.