将线段AB延长到C,使BC=1/3AB,反向延长AC到D点使AD=1/2AC,则CD=多少AB?

问题描述:

将线段AB延长到C,使BC=1/3AB,反向延长AC到D点使AD=1/2AC,则CD=多少AB?

BC=1/3AB
AC=AB+BC=4/3AB
AD=1/2AC=(1/2)×(4/3AB)=2/3AB
CD=AD+AC
=2/3AB+2/3AB
=(2/3+4/3)AB
=2AB