作DE⊥AB于E,DF⊥BC于F.
又∵∠ABC=90°,
则四边形EBFD为矩形.
∵BD平分∠ABC,
∴DE=DF,
∴矩形EBFD为正方形.
∴DE=BE=BF=DF,
RTΔCFD和RTΔCBA中,
∠DCF=∠ACB,
∴RTΔCFD∽RTΔCBA,
有:
DF/AB=CF/BC,CD/AC=DF/AB;
∴DF/AB=(BC-BF)/BC=(BC-DF)/BC,CD=DF*AC/AB;
∴DF/6=(8-DF)/8,CD=DF*10/6;
∴DF=24/7,CD=(24/7)(10/6)=40/7;
即CD=40/7.