#### 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.

#### Solution

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.

Is there an error in this question or solution?

Solution 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. Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.