Question

如图3，在三角形ABC中，AD平分∠BAC,过B作BE垂直AD，垂足为点E，过E作EF//AC交AB于F，AF与BF关系

Answer 1

∠AEB＝90º, ∠FEA＝∠EAC＝∠EAB, ∴ AF＝FE.

∠ABE＝90°-∠BAE＝90°-∠AEF＝∠FEB. ∴FE＝FB, AF＝BF

Answer 2

AF=BF
AD平分∠BAC,∠BAD=∠CAD.EF//AC,∠CAD=∠AEF
∠BAD=∠AEF,AF=DF
BE垂直AD,∠BEF+∠AEF=90,即∠BEF+∠BAD=90.
∠ABE+∠BAD=90,∠BEF=∠ABE,BF=DF
AF=BF