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.