解：(1)$∵∠ DAE=∠ BAC$,
$∴∠ DAE-∠ BAE=∠ BAC-∠ BAE$,
$∴∠ BAD=∠ CAE$,
又$∵\frac {AD}{AE}=\frac {AB}{AC}=\frac {4}{5}$,
$∴△ BAD ∽ △ CAE$,
$∴\frac {BD}{CE}=\frac {AD}{AE}=\frac {4}{5}$
(2)如图所示，过点$B$作$BG⊥ AD$交$AD$延长线于$G$，设$DF$，$AC$交于$H$，
在$Rt△ ABC$中，$∠ ACB=90°$，$\sin ∠ BAC=\frac {BC}{AC}=\frac {3}{5}$,
设$BC=3y$，$AB=5y$，则$AC=\sqrt {AB^2-BC^2}=4y$,
$∵∠ AED=90°$，$∠ DAE=∠ BAC$,
$∴\sin ∠ DAE=\sin ∠ BAC=\frac {3}{5}$,
$∴\frac {DE}{AD}=\frac {3}{5}$,
设$AD=5x$，$DE=3x$，则$AE=\sqrt {AD^2-DE^2}=4x$
设$BD=10m$,
$∵∠ ADB=120°$,
$∴∠ BDG=60°$,
$∴DG=BD· \cos ∠ BDG=5m$，$BG=BD· \sin ∠ BDG=5\sqrt {3}m$,
同理可证明$∠ CAE=∠ BAD$,
又$∵∠ AEH=180°-∠ AED=90°=∠ G$,
$∴△ AEH ∽ △ AGB$,
$∴\frac {HE}{BG}=\frac {AH}{AB}=\frac {AE}{AG}$，
即$\frac {HE}{5\sqrt {3}m}=\frac {AH}{5y}=\frac {4x}{5x+5m}$,
$∴AH=\frac {4xy}{x+m}$，$EH=\frac {4\sqrt {3}mx}{x+m}$,
$∴CH=AC-AH=\frac {4my}{x+m}$
$∵∠ HCF=180°-∠ ACB=90°=∠ HEA$,
$∠ CHF=∠ EHA$,
$∴△ CHF ∽ △ EHA$,
$∴\frac {CF}{AE}=\frac {CH}{HE}$，
即$\frac {CF}{4x}=\frac {\frac {4my}{x+m}}{\frac {4\sqrt {3}mx}{x+m}}$,
$∴CF=\frac {4\sqrt {3}y}{3}$,
$∴\frac {BC}{CF}=\frac {3y}{\frac {4\sqrt {3}y}{3}}=\frac {3\sqrt {3}}{4}$
