Two circles touch each other internally at a point $P$ . A chord $A B$ of the bigger circle intersects the other circle in $C$ and $D$ .Prove that : $∠ C P A = ∠ D P B$

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Solution

Draw a tangent TS at P to the circles given
Since TPS is the tangent, PD is the chord.
$∴ ∠ P A B = ∠ B P S . . . . (1)$ [Angles in alternate segment]
Similarly,
$∠ P C D = ∠ D P S . . . . (2)$
Subtracting (1) and (2)
$∠ P C D - ∠ P A B = ∠ D P S - ∠ B P S$
But in $△ P A C$
Exterior $∠ P C D = ∠ P A B + ∠ C P A$
$∴ ∠ P A B + ∠ C P A - ∠ P A B = ∠ D P S - ∠ B P S$
$\Rightarrow ∠ C P A = ∠ D P B$

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