Question

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

Answer 1

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[ refer above attachment. ]

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]

[ Exterior angle of a triangle is equal to sum of interior opposite angles. ]

Hence, in ΔAPT, ∠APO = ∠a + ∠b

In ΔAPO, OA = OP [Radii of circle]

∠OAP = ∠OPA = ∠a + ∠b ---(i) [Angles opposite to equal sides are equal]

From the figure, ∠OAP=90°- ∠b ----(ii) [Since radius is perpendicular to tangent]

From equations (i) and (ii),

∠a + ∠b = 90° - ∠b

⇒ ∠a = 90° - 2∠b ---(iii)

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 = ∠c

Hence,
AP bisects ∠TAB

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Hope it helps! :D