延长AB,AM交于E;延长CM至F使MF=CM;连接DE;∵CD平分∠ACB,AM⊥CD于M;∴AC=EC;AD=DE;∠FDE=∠ADF=∠CDB;AM=EM;∵CD=BC;∴∠CBD=∠CDB=∠FDE;∵MF=CM;AM⊥CD∴CE=FE;∠DEF=∠MEF+∠DEM=∠CEM+∠BAE=∠ABC=∠FDE;∴DF=FE;;∵MF=CM;AM=EM;∴△AMC≌△EMF;AC=EF;∴MF=CM;FC=CD+DF=2*CM;CM=1/2(CD+DF)=1/2(BC+DF);=1/2(BC+EF)=1/2(BC+AC);得证;
如图,△ABC中,CD平分∠ACB,AM⊥CD于M,CD=CB,求证CM=1/2(AC+BC)
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