由AD⊥BC,∴∠ADB=90º,
BE⊥AC,∴∠AEB=90º,
△AEF与△BDF中,(其中F是AD,BE交点)
∴∠EAF=∠DBF,
∴△EAF∽△DBF
∴EF/AF=DF/BF,∠EFD=∠AFB,
∴△EFD∽△AFB
ED/AB=EF/AF
又△AEF∽△ADC
∴EF/AF=CD/AC=3/5
∴DE/AB=3/5.
由AD⊥BC,∴∠ADB=90º,
BE⊥AC,∴∠AEB=90º,
△AEF与△BDF中,(其中F是AD,BE交点)
∴∠EAF=∠DBF,
∴△EAF∽△DBF
∴EF/AF=DF/BF,∠EFD=∠AFB,
∴△EFD∽△AFB
ED/AB=EF/AF
又△AEF∽△ADC
∴EF/AF=CD/AC=3/5
∴DE/AB=3/5.