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Answer» Hola!
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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
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