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春芽
共回答了15个问题采纳率:93.3% 举报
证明:FD延长线,取g GD = FD连接EG BG
∵∠A = 90
∴∠ABC +∠C = 90
∵DBC的中点
∴BD = CD
∵GD = FD,∠BDG =∠CDF
∴△BDG≌△CDF(SAS),
∴BG = CF∠GBD =∠C
∴ ∠ABG =∠ABC +∠GBD =∠ABC +∠C = 90
∴EG 2 = 2 + BG 2
∴EG 2 = 2 + CF 2
∵DE⊥DF ,GD = FD
∴DE垂直平分线FG
∴EF = EG
∴EF 2 = 2 + CF 2
1年前
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