(1)证明：∵AF平分∠BAC，∴∠CAD= ∠DAB=∠BAC． ∵点D与点A关于点E对称∴E为AD中点． BC⊥AD，∴BC为AD的中垂线，∴AC= CD∵在Rt△ACE和Rt△ABE中，∠CAD+∠ACE=∠DAB+ ∠ABE =．∠CAD= ∠DAB.∴∠ACE= ∠ABE，∴AC =AB，∴AB= CD. (2)∵∠BAC =2∠MPC，又∵∠BAC =2∠CAD，∴∠MPC=∠CAD．∵AC= CD，∴∠CAD=∠CDA，∵∠MPC=∠CDA.∴∠MPF=∠CDM，∵AC =AB，AE⊥BC，∴CE= BE，∴AM为BC的中垂线，∴CM=BM∵EM⊥BC，∴EM平分∠CMB，（等腰三角形三线合一）∴∠CME=∠BME．∴∠BME= ∠PMF，∠PMF=∠CME， ∴∠MCD= ∠F(三角形内角和)．