$解：(1)∵AC平分∠BAD$
$∴∠BAC=∠CAD$
$∵BC⊥AC，CD⊥AD$
$∴∠ACB=∠ADC=90°$
$∴△ABC∽△ACD$
$∴\frac {AB}{AC}=\frac {AC}{AD}$
$∵AB=18，AC=12$
$∴\frac {18}{12}=\frac {12}{AD}$
$∴AD=8$
$(2)∵△ABC∽△ACD$
$∴\frac {DE}{CF}=\frac {AC}{AB}=\frac 23$
