∠∠ACD =∠BAC+∠ABC

∴ 1/2∠acd= 1/2∠bac+ 1/2∠abc

∠∠PBC = 1/2∠ABC∠PCD = 1/2∠ACD

∴∠PCD=∠ 1/2∠BAC+∠PBC

∠∠PCD =∠PBC+∠BPC

∴ 1/2∠BAC=∠PBC

∴∠BAC=2∠PBC=80

∴∠CAE= 180 -∠BAC= 100

Let PF⊥AB be in F, PM⊥BC be in M, and PN⊥AC be in N.

∠ABP=∠CBP ∠DCP=∠ACP

∴PF=PM PM=PN (the distance between two points on the bisector of an angle is equal to both sides of the angle)

∴PF=PN

∴∠ Cap =1/2∠∠∠ CAE = 50 (points with equal distance to both sides of the angle are on the bisector of this angle).

2. solution: AD=BE+DE

∠∠ADC =∠ACB = 90°

∴∠DAC+∠DCA=∠DCA+∠ECB=90

∴∠DAC=∠ECB

∠∠ADC =∠CEB = 90∠DAC =∠ECB AC = CB

∴⊿ADC≌⊿CEB

∴AD=CE CD=BE

∫CE = CD+DE

∴AD=BE+DE