2014年临沂市中考数学试卷及答案word版(12)

N （1）（本小问4分） 解不等式，得v≥

方法一：过点E作EF⊥AM，垂足为F.

∵AE平分∠DAM，ED⊥AD，

E ∴ED=EF. ··········································· （1分）

由勾股定理可得,

AD=AF. ··············································· （2分）

又∵E是CD边的中点， G B M C ∴EC=ED=EF. 又∵EM=EM， ∴Rt△EFM≌Rt△ECM. ∴MC=MF. ························································· ····················································· （3分） ∵AM=AF+FM， ∴AM=AD+MC. ······································································································· （4分） 方法二：

连接FC. 由方法一知，∠EFM=90°, AD=AF，EC=EF. ······································· （2分） 则∠EFC=∠ECF， ∴∠MFC=∠MCF. ∴MF=MC. ··············································································································· （3分） ∵AM=AF+FM， ∴AM=AD+MC. ······································································································· （4分） 方法三：

延长AE，BC交于点G. ∵∠AED=∠GEC，∠ADE=∠GCE=90°，DE=EC， ∴△ADE≌△GCE. ∴AD=GC, ∠DAE=∠G. ··························································································· （2分） 又∵AE平分∠DAM， ∴∠DAE=∠MAE， ∴∠G=∠MAE， ∴AM=GM， ············································································································ （3分）

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