题目内容
(11·丹东)(本题12分)已知:正方形ABCD.
(1)如图1,点E、点F分别在边A
B和AD上,且AE=AF.此时,线段BE、DF的数量关
系和位置关系分别是什么?请直接写出结论.
(2)如图2,等腰直角三角形FAE绕直角顶点A顺时针旋转
,当
时,连接BE、DF,此时(1)中结论是否成立,如果成立,请证明;如果不成立,请说明理由.
(3)如图3,等腰直角三
角形FAE绕直角顶点A顺时针旋转
,当
时,连接BE、DF,猜想当AE与AD满足什么数量关系时,直线DF垂直平分BE.请直接写出结论.
(4)如图4,等腰直角三角形FAE绕直角顶点A顺时针旋转
,当
时,连接BD、DE、EF、FB得到四边形BDEF,则顺次连接四边形BDEF各边中点所组成的四边形是什么特殊四边形?请直接写出结论.
(1)如图1,点E、点F分别在边A


(2)如图2,等腰直角三角形FAE绕直角顶点A顺时针旋转


(3)如图3,等腰直角三



(4)如图4,等腰直角三角形FAE绕直角顶点A顺时针旋转



(1)BE=DF且BE⊥DF
(2)成立
证明:延长DF交AB于点H,交BE于点G.

在
中,
∵
∴
∴
又∵
∴
∴BE=DF且BE⊥DF仍成立
(3)
(4)菱形
(2)成立
证明:延长DF交AB于点H,交BE于点G.

在

∵

∴

∴

又∵

∴

∴BE=DF且BE⊥DF仍成立
(3)

(4)菱形
略

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