如图,在△ABC中,AD平分∠BAC,DF⊥AB于点F,E为AC上一点,且AE=DE

1个回答

  • (1)∵AE=DE∴∠EAD=∠EDA∵AD平分∠BAC∴∠EAD=∠BAD∴∠BAD=∠EDA∴AB//DE∵DF⊥AB ∴DF⊥DE∴∠EDF=90°∵DF=2,DE=AE=3∴EF=√13(根据勾股定理)(2)证明:作DG//AC,交AB于G∵AB//DE∴四边形AEDG是平行四边形∵AE=DE∴四边形AEDG是菱形(邻边相等的平行四边形是菱形)∴AG=AE∵AB//DE∴∠BAC+∠AED=180°∵∠B+∠AED=180°∴∠B=∠BAC∵DG//AC∴∠BGD=∠BAC∴∠B=∠BGD∴BD=GD∵DF⊥AB∴BF=FG(等腰三角形三线合一)∵BF=AB-AF FG=AF-AG=AF-AE∴AB-AF=AF-AE∴AB+AE=2AF