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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.
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