1）∵∠C=90°,則tanB=AC/BC=3/4;又AB=5.設(shè)AC=3X
∴AC^2+BC^2=AB^2,即25X^2=25,X=1.則AC=3,BC=4.
當(dāng)EF∥BC時(shí),則△AFE∽△ACB.故AF:FE:EA=AC:CB:BQ=3:4:5.
設(shè)FE=4m,由∠CDF=∠DFE；∠C=∠FDE=90°可知：△FCD∽△EDF,
則DF/EF=CD/DF,DF^2=CD*EF=2EF=8m.
∵DF^2-CD^2=CF^2,即8m-2^2=(3-3m)^2
∴m=(13-2√13)/9[(m=13+2√13)/9不合題意,舍去）
則BE=AB-AE=5-5m=5-5*(13-2√13)/9=(10√13-20)/9.
2)當(dāng)點(diǎn)E在CB的中垂線上即ED⊥CB、F與C重合時(shí),△EDC≌△EDB∽
△ACB.則BE/BA=BD/BC,BE/5=2/4,則BE=2.5.弟二小題應(yīng)該是△DEF∽△CAB∽△DEB 吧~~~自己明白就好