（见图）延长CM、AB交于E,作EF//BC并与AD延长线交于F,由此得AE=AC(由AD平分∠BAC,CM⊥AD,得三角形AEC为等边三角形）BE=DF（由AB=AD,BD//EF,得AE=AF)由三角形AEC是等边三角形,AM⊥EC,得EM=MC由BC//EF.得∠CDM=∠EFM由...