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.

Answer 1

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