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Question
Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that ∠CPA = ∠DPB.
Solution
Draw a tangent TS at P to the circles given.
Since TPS is the tangent, PD is the chord.
∴ ∠PAB = ∠BPS ...(i) (Angles in alternate segment)
Similarly,
∠PCD = ∠DPS ...(ii)
Subtracting (i) from (ii)
∠PCD – ∠PAB = ∠DPS – ∠BPS
But in ∠PAC,
Ext. ∠PCD = ∠PAB + ∠CPA
∴ ∠PAB + ∠CPA – ∠PAB = ∠DPS – ∠BPS
`=>` ∠CPA = ∠DPB
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