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.
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)
`∠`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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- Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers