Mathematics
In the given figure, PAB is a secant to the circle and PT is a tangent at T. Prove that:
(i) ∠PAT ∼ ∠PTB
(ii) PA × PB = PT2
Circles
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Answer
(i) In △PAT and △PTB:
⇒ ∠TPA = ∠BPT [Common angles]
⇒ ∠PTA = ∠PBT [Angles in the alternate segment]
△PAT ∼ △PTB [By AA similarity]
Hence, proved △PAT ∼ △PTB.
(ii) Since, corresponding sides of similar triangle are proportional to each other.
PT2 = PA × PB
Hence, proved that PT2 = PA × PB.
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