（本題7分）如圖，在Rt△ABC中，∠C=90°，點D是CB的中點，將△ACD沿AD摺疊後得到△AED，過點B...

問題詳情：

（本題7分）如圖，在Rt△ABC中，∠C=90°，點D是CB的中點，將△ACD沿AD摺疊後得到△AED，過點B作BF∥AC交AE的延長線於點F. 求*：BF=EF.

【回答】

*：如答圖，連接DF

∵D 是CB的中點，

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

∵將△ACD沿AD摺疊後得到△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

知識點：未分類

題型：未分類