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Two circles are internally tangent at $T$. $AB$ is a chord of the outer circle that is also tangent to the inner circle at $P$.

How can one go about showing that $\angle ATP = \angle BTP$?

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1 Answer 1

up vote 1 down vote accepted

I found a proof.

Draw the tangent at $T$ and label two points on it $C$ and $D$ with $C$ on the "$A$" side and $D$ on the other side of $T$. Let $AT$ intersect with the smaller circle at S.

Then

$\angle ATC = \angle TPS$

$\angle BTD = \angle BAT$

$\angle APS = \angle PTA$

and the conclusion follows.

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