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