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.
