#### 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

Is there an error in this question or solution?

Solution 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. Concept: Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers.