TA and TB are tangents to a circle withO from an external point T. OTcircle at point P. Prove that AP bisects the angleangle TABcentre

TA and TB are tangents to a circle withO from an external point T. OTc

1.	TA and TB are tangents to a circle withO from an external point T. OTcircle at point P. Prove that AP bisects the angleangle TABcentre
Answer» GivenTA and TB are tangent to a circle with centre O from point TConsider, ΔOAT and ΔOBTHere, 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]Recall that exterior angle of a triangle is equal to sum of interior opposite angles.Hence in ΔAPT, ∠APO = ∠a + ∠bIn ΔAPO, OA = OP [Radii of circle]∠OAP = ∠OPA = ∠a + ∠b ... (1) [Angles opposite to equal sides are equal]From the figure, ∠OAP=90°- ∠b ... (2) [Since radius is perpendicular to tangent]From equations (1) and (2), we get∠a + ∠b = 90° - ∠b⇒ ∠a = 90° - 2∠b … (3)In ΔOAT, ÐOAT + ÐAOT + ÐATO = 180° [Angle sum property of triangle]90° + ∠a + ∠b+∠c+=180°∠a+∠b+∠c=90°90°- 2∠b + ∠b + ∠c=90°- ∠b + ∠c=0∠b = ∠cHence, AP bisects ∠TAB