$解:作DG//AF{交}BC于点G$
$∵DG//AF$
$∴\frac {AD}{CD}=\frac {FG}{CG},\frac {BE}{ED}=\frac {BF}{FG}$
$∵\frac {AD}{CD}=\frac 23,\frac {BE}{ED}=\frac 32$
$∴\frac {FG}{CG}=\frac 23,\frac {BF}{FG}=\frac 32$
$设FG=2x,则CG=3x,BF=3x$
$∴FC=FG+CG=5x$
$∴BF:FC=3:5$