已知AE⊥AB,AF⊥AC,AE=AB,AF=AC。试证明:(1)EC=BF;(2)EC⊥BF

显示全部楼层 · 2012-9-23 12:47:29

画图：（1）∵AE=AB，AF=AC而∠EAC = 90°+∠BAC，∠BAF= 90°+∠BAC，即∠EAC = ∠BAF∴△AEC ≌△ABF（SAS）∴EC = BF(2)设BF角CE于O∵△AEC ≌△ABF (已证)∴∠AFB = ∠ACE而AF⊥AC，AF=AC∴△ACF是直角等腰三角形，∠AFC = ∠ACF = 45°又∵在△CFO中，∠OFC = 45° -∠AFB，∠OCF = 45° +∠AFB∴∠OFC 与∠OCF互余， ∠COF = 90°∴EC⊥BF...

千问 · 2012-9-23 12:47:29

1）∵AE=AB，AF=AC而∠EAC = 90°+∠BAC，∠BAF= 90°+∠BAC，即∠EAC = ∠BAF∴△AEC ≌△ABF（SAS）∴EC = BF(2)设BF角CE于O∵△AEC ≌△ABF (已证)∴∠AFB = ∠ACE而AF⊥AC，AF=AC∴△ACF是直角等腰三角形，∠AFC = ∠ACF =...