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 tangent at point P,

∠PAB = ∠BPS ………….(1) [Angles in alternate segments are equal.]

∠PCD = ∠DPS ………….(2) [Angles in alternate segments are equal.]

Subtracting (1) from (2), we get :

⇒ ∠PCD - ∠PAB = ∠DPS - ∠BPS …………(3)

In △PAC,

⇒ ∠PCD = ∠PAC + ∠CPA [An exterior angle is equal to sum of two opposite interior angles.]

⇒ ∠PCD = ∠PAB + ∠CPA [From figure, ∠PAC = ∠PAB]

Substituting above value of ∠PCD in (3), we get :

⇒ ∠PAB + ∠CPA - ∠PAB = ∠DPS - ∠BPS

⇒ ∠CPA = ∠DPB [From figure, ∠DPS - ∠BPS = ∠DPB].

Hence, proved that ∠CPA = ∠DPB.