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.
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.
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
