证明：作DG⊥AC,DH⊥AB,垂足分别是E、F.∵AD是∠BAC的角平分线DG⊥AC,DH⊥AB∴DE=DF∵∠AED＋∠AFD=180°.且∠AED＋∠HED=180°.∴∠HED=∠AFD∵DG⊥AC,DH⊥AB,∴∠EHD=∠DGF=90°∴⊿DHE≌⊿DGF∴DE=DF(2)∵DG⊥AC,...