延长CM交AB延长线于F，取BF中点E，联接ME∵AM⊥CF ∠CAM=∠FAM∴∠ACF=∠F∴AC=AF∴CM=FM∵BE=EF∴EM∥BC∴AD/AM=AB/AE∵AD=AB∴AM=AE∵AC=AF=AE+EF AB=AE-BE=AE-EF∴AB+AC=2AE=2AD∴AD=1/2(AB+AC)
