∵∠C=90°,AC=8,BC=6∴勾股定理：AB=10做DE⊥AB于E∵AD平分∠BAC,DE⊥AB,DC⊥AC(∠ACD=∠ACB=90°)∴CD=DES△ACD+S△ABD=S△ABC1/2CD×AC+1/2DE×AB=1/2×AC×BCCD(AC+AB)=AC×BCCD(8+10)=8×6CD=8/3...