Given:
- Two circles touch each other internally at point P.
- Chord AB of the larger circle intersects the smaller circle at points C and D.

To prove:
- ∠APC = ∠BPD

Proof:
Step 1: Connect the centers of the circles.
- Let O1 be the center of the larger circle and O2 be the center of the smaller circle.
- Draw lines OP1 and OP2, where O1P1 and O2P2 are the radii of their respective circles.

Step 2: Draw perpendiculars from P to the chord AB.
- Let the perpendiculars intersect AB at points M and N, where M is on the larger circle and N is on the smaller circle.

Step 3: Observe the congruent triangles.
- Triangles O1PM and O2PN are congruent by the hypotenuse-leg congruence criterion.
- ∠O1PM = ∠O2PN (corresponding angles of congruent triangles).

Step 4: Prove the congruent triangles.
- O1P is the radius of the larger circle, and O2P is the radius of the smaller circle.
- PM and PN are perpendiculars to AB, which is a chord of both circles.
- Therefore, O1PM and O2PN are right triangles.

Step 5: Prove the congruent right triangles.
- Both triangles have a right angle at P.
- Both triangles have a common side OP, which is the radii of the circles.
- Both triangles have another common side PM and PN, which are perpendiculars from P to the chord AB.

Step 6: Prove the congruent angles.
- Since the triangles are congruent, their corresponding angles are also congruent.
- ∠O1PM = ∠O2PN.

Step 7: Prove the required angles.
- ∠APC = ∠O1PM (corresponding angles)
- ∠BPD = ∠O2PN (corresponding angles)
- Since ∠O1PM = ∠O2PN (proved in Step 6), we can conclude that ∠APC = ∠BPD.

Therefore, we have proved that ∠APC = ∠BPD.