题目内容
如图,直角梯形ABCD中,AD∥BC,∠A=90°,AB=AD=6,DE⊥DC交AB于E,DF平分∠EDC交BC于F,连接EF.

(1) 证明:EF=CF;
(2) 当AE=2时,求EF的长.

(1) 证明:EF=CF;
(2) 当AE=2时,求EF的长.
(1)见解析, (2)EF = 5
解:(1) 如图,过D作DG⊥BC于G
由已知可得四边形ABGD为正方形
∵DE⊥DC
∴∠ADE+∠EDG=90°=∠GDC+∠EDG
∴∠ADE=∠GDC
在△ADE与△GDC中,
∴△ADE≌△GDC (ASA) ···························· 3分
∴DE=DC且AE=GC
在△EDF和△CDF中
∴△EDF≌△CDF(SAS)··························· ·6分
∴EF=CF··································· 7分
(2) ∵AE=2
设EF=x,则BF=8-CF=8-x,BE=4
由勾股定理x2=
+42
解得
∴EF = 5 12分
(1)过D作DG⊥BC于G,可得四边形ABGD为正方形,求得△ADE≌△GDC (ASA),△EDF≌△CDF(SAS),从而得出结论
(2)利用勾股定理求解
由已知可得四边形ABGD为正方形
∵DE⊥DC
∴∠ADE+∠EDG=90°=∠GDC+∠EDG
∴∠ADE=∠GDC
在△ADE与△GDC中,

∴△ADE≌△GDC (ASA) ···························· 3分
∴DE=DC且AE=GC
在△EDF和△CDF中

∴△EDF≌△CDF(SAS)··························· ·6分
∴EF=CF··································· 7分
(2) ∵AE=2
设EF=x,则BF=8-CF=8-x,BE=4
由勾股定理x2=

解得

∴EF = 5 12分
(1)过D作DG⊥BC于G,可得四边形ABGD为正方形,求得△ADE≌△GDC (ASA),△EDF≌△CDF(SAS),从而得出结论
(2)利用勾股定理求解

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