问题描述：
如图,在△ABC中,AD平分∠BAC,AD=AB,CM⊥AD于M.求证:AM=二分之一(AB+AC)
大家来看看,急
能够延长AB,CM交于P,取BP的中点N,连接MN，
大家来看看,急
能够延长AB,CM交于P,取BP的中点N,连接MN，
问题解答：
我来补答
证明：
延长AM至E 使得AE=AC,连结EC
∵AD平分∠BAC
∴∠BAD=∠CAE
∵AB=AD ,AE = AC
∴△ABD∽△AEC
∴AB/AC=BD/EC
∵∠BAD = ∠CAD,AB = AD,AC = AE
∴2∠AEC = 2∠B = 2∠ADB = 2∠CDE
∴∠AEC = ∠CDE
∴CD = CE
∵CM⊥DE
∴DM=EM
∴AM
=AD+DM
=(AD+AD+DM+DM)/2
=(AD+AD+MD+ME)/2
=(AD+AE)/2
=(AD+AC)/2
=(AB+AC)/2
延长AM至E 使得AE=AC,连结EC
∵AD平分∠BAC
∴∠BAD=∠CAE
∵AB=AD ,AE = AC
∴△ABD∽△AEC
∴AB/AC=BD/EC
∵∠BAD = ∠CAD,AB = AD,AC = AE
∴2∠AEC = 2∠B = 2∠ADB = 2∠CDE
∴∠AEC = ∠CDE
∴CD = CE
∵CM⊥DE
∴DM=EM
∴AM
=AD+DM
=(AD+AD+DM+DM)/2
=(AD+AD+MD+ME)/2
=(AD+AE)/2
=(AD+AC)/2
=(AB+AC)/2
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