题目

如图，∠ABC=90°，D、E分别在BC、AC上，AD⊥DE，且AD=DE. 点F是AE的中点，FD的延长线与AB的延长线相交于点M，连接MC. (1)求证：∠FMC=∠FCM； (2)AD与MC垂直吗？说明你的理由. 答案：解：（1）证明：∵△ADE是等腰直角三角形，F是AE的中点. ∴DF⊥AE，DF=AF=EF. ··············································································· 1分 又∵∠ABC=90°，∠DCF、∠AMF都与∠MAC互余， ∴∠DCF=∠AMF. ························································································· 2分 又∵∠DFC=∠AFM=90°， ∴△DFC≌△AFM（ASA）. ··········································································· 3分 ∴CF=MF. ····································································································· 4分 ∴∠FMC=∠FCM. ························································································· 5分 （2）AD⊥MC. 理由如下： 如图，延长AD交MC于点G. 由（1）知∠MFC=90°，FD=FE，FM=FC. ∴∠FDE=∠FMC=45°，·················································································· 6分 ∴DE//CM. ····································································································· 7分 ∴∠AGC=∠ADE=90°，·················································································· 8分 ∴AG⊥MC，即AD⊥MC. ·············································································· 9分

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