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证明:过T作TF⊥AB于F,
∵AT平分∠BAC,∠ACB=90°,
∴CT=TF(角平分线上的点到角两边的距离相等),
∵∠ACB=90°,CM⊥AB,
∴∠ADM+∠DAM=90°,∠ATC+∠CAT=90°,
∵AT平分∠BAC,
∴∠DAM=∠CAT,
∴∠ADM=∠ATC,
∴∠CDT=∠CTD,
∴CD=CT,
又∵CT=TF(已证),
∴CD=TF,
∵CM⊥AB,DE∥AB,
∴∠CDE=90°,∠B=∠DEC,
在△CDE和△TFB中,
| ∠B=∠DEC | ∠CDE=∠TFB=90° | CD=TF |
| |
,
∴△CDE≌△TFB(AAS),
∴CE=TB,
∴CE-TE=TB-TE,
即CT=BE.