连接AE,
∵DE垂直平分AB,∴AE=BE,
设AE=BE=X,则CE=10-X,
在RTΔACE中,AE^2=AC^2+CE^2,
X^2=36+(10-X)^2,X=34/5,
∴BE=34/5,CE=10-34/5=16/5.