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

Given TA and TB are tangent to a circle with centre O from point T.

Consider, ΔOAT and ΔOBT

Here, OA = OB [Radii of circle]

OT = OT [Common side]

TA = TB [Tangents from an external point to a circle are equal in length]

∴ ΔOAT ≅ ΔOBT [By SSS congruence criterion]

∴ ∠ATO = ∠BTO [By C.P.C.T.]

⇒ ∠ATP = ∠BTP [From figure, ∠ATO = ∠ATP and ∠BTO = ∠BTP]

In ΔAPT and ΔBPT,

AT = BT [Tangents from an external point to a circle are equal in length]

PT = PT [Common side]

∠ATP = ∠BTP [Proved above]

∴ ΔAPT ≅ ΔBPT [By SAS congruence criterion]

∴ ∠PAT = ∠PBT [By C.P.C.T.]

and AP = BP [By C.P.C.T.]

In ΔPAB,

∠PAB = ∠PBA [Angles opposite to equal sides are equal]

∠PAT = ∠PBA [Angles in alternate segments are equal]

∴ ∠PAB = ∠PAT

∴ AP is the bisector of ∠TAB

Hence, proved that AP bisects ∠TAB.