Prove:

Because BD divides ∠ABC equally,

So ∠ PBC = ∠ ABC/2

Similarly ∠ PCB = ∠ ACB/2

Because PBC+PCB+BPC = 180.

So ∠ BPC = 180-∠ PBC-∠ PCB.

= 180-(∠ABC+∠ACB)/2

Because ∠ A+∠ ABC+∠ ACB = 180.

So ∠ ABC+∠ ACB =180-∠ A.

so∠BPC = 180-( 180-∠A)/2

=90 + 1/2∠A

The second picture:

Prove:

According to the internal angle and nature of the triangle:

∠ABC+∠ACB= 180 -∠A

∠BOC= 180 -(∠CBO+∠BCO)

Because BO and CO are the bisector of ∠ABC, the outer corner of ACB, DBC and ECB.

So ∠ DBO = ∠ CBO = ∠ CBD/2

ECO = ecology =∠BCE/2

So ∠ BOC = 180-(∠ CBO+∠ BCO)

= 180 -(∠CBD/2+∠BCE/2)

= 180 -(∠CBD+∠BCE)/2

Because ∠ CBD = 180-∠ ABC, ∠ BCE = 180-∠ ACB.

so∠BOC = 180-( 180-∠ABC+ 180-∠ACB)/2

=(∠ABC+∠ACB)/2

=( 180 -∠A)/2

So ∠ BOC = 90-1/2 ∠ A.

The third picture:

Proof: ∫2∠PCM = 2∠PBC+∠A,?

∴∠? PCM？ =∠PBC？ + 1/2∠A，？

∵∠PCM？ =∠PBC？ +∠BPC？

∴∠BPC= 1/2∠A。