（本题7分）如图，在Rt△ABC中，∠C=90°，点D是CB的中点，将△ACD沿AD折叠后得到△AED，过点B...

问题详情：

（本题7分）如图，在Rt△ABC中，∠C=90°，点D是CB的中点，将△ACD沿AD折叠后得到△AED，过点B作BF∥AC交AE的延长线于点F. 求*：BF=EF.

【回答】

*：如答图，连接DF

∵D 是CB的中点，

 ∴CD=BD. ····················································································· 1 分

∵将△ACD沿AD折叠后得到△AED，

∴CD=ED，∠AED=∠C=90°. ·······························································2 分

 ∴BD=ED，∠DEF =90°. ······································································ 3 分

∵BF∥AC，∠C=90°，

∴∠CBF=90°.

∴∠DBF=∠DEF=90°. ·········································································4 分

 在 Rt△DBF 和 Rt△DEF 中，

∴Rt△DBF≌Rt△DEF（HL）.···································································6 分

∴BF=EF. ······················································································ 7 分

知识点：未分类

题型：未分类