∵AC=1/3BC
∴AC=1/4AB,BC=3/4AB
∵D为BC中点
∴CD=1/2BC=3/8AB
AD=AC+CD=5/8AB
∵E为AD中点
∴AE=1/2AD=5/16AB
CE=AE-AC=1/16AB
∴AB=16CE=16*105=1680
3AC/2=CD,(CD+AC)/2=AE,AE-AC=CE.
=>(CD+AC)/2-AC=CE
=>(3/2AC+AC)/2-AC=CE
解得 AC 。AB=4AC
设AB长为x
则AC长为x/4
AD长为AC+BC/2=x/4+(x-x/4)/2=5x/8
AE长为5x/8/2=5x/16
CE长为AE-AC=5x/16-x/4=105cm
x=1680cm