First of all, we can find the degree of the angle BAC: from the triangle external angle theorem, there are: ∠ PCD = ∠ PBC+∠ BPC.

∵∠PBC=∠ABC/2、∠pcd=∠acd/2,∴∠acd/2=∠abc/2+∠bpc，

∴∠ACD=∠ABC+2∠BPC。 ①

From the triangle exterior angle theorem, there are: ∠ ACD = ∠ ABC+∠ BAC. ②

Comparing ① and ②, we get: ∠ BAC = 2 ∠ BPC = 2× 40 = 80.

Secondly, we can start looking for ∠APC times: extend BA to E, so that AE = AB.

∠∠ABC =∠ACB,∠BAC = 80 ,∴∠abc=50 ,∴∠pbc=25 .

∠∠ABC =∠ACB, ∴ AB = AC, while AE = AB, ∴ AB = AC = AE, ∴ BEC = ∠ ACE.

According to the triangle exterior angle theorem, there are: ∠ BAC = ∠ BEC+∠ ACE = 2 ∠ BEC, ∴∠ BEC = ∠ BAC/2 = 40.

From ∠ BEC = 40 and ∠ BPC = 40, we get: B, E, P, C*** circle, AB = AC = AE.

∴A is the center of the circumscribed circle of the quadrilateral BEDC, ∴∠ Cap = 2 ∠ PBC = 50.