In figure, two circles touch internally at a point P. AB is a chord of the bigger circle touching the other circle at C. Prove that PC bisects the angle APB.[Hint: Draw a tangent at the point P. Joint CD, where D is the point of intersection of AP and the inner circle and prove that ∠ PBC = ∠ PCD.]

Given: Two circles touch internally at a point P. AB is chord of the bigger circle touching the other circle at C.
To prove: ∠ BPC = ∠ CPA

Proof:
In Δ BCP and Δ CDP,
∠ XPA = ∠ PBA (Alternate segment angles)
∠ PCB = ∠ CDP (Alternate segment angles)
Δ BCP ≅ Δ CDP (By AA similarity)
∠ BPC = ∠ CPD
∴ ∠ BPC = ∠ CPA
∴ PC bisect ∠ BPA.