af平分角bac

(1)證明：∵AF平分∠BAC，∴∠CAD= ∠DAB=∠BAC． ∵點D與點A關於點E對稱∴E爲AD中點． BC⊥AD，∴BC爲AD的中垂線，∴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爲BC的中垂線，∴CM=BM∵EM⊥BC，∴EM平分∠CMB，（等腰三角形三線合一）∴∠CME=∠BME．∴∠BME= ∠PMF，∠PMF=∠CME， ∴∠MCD= ∠F(三角形內角和)．