证明:∵$AD$平分$∠BAC,$∴$∠CAD = ∠EAD$
∵$DE\perp AB,$∴$∠AED = 90°$
又∵$∠ACB = 90°,$∴$∠ACD = ∠AED$
在$\triangle ACD$和$\triangle AED$中
$\begin {cases}∠ACD = ∠AED\\∠CAD = ∠EAD\\AD = AD\end {cases}$
∴$\triangle ACD≌\triangle AED(\mathrm {AAS})$
由全等可得$AC = AE,$$CD = ED$
∴点$A,$$D$都在线段$CE$的垂直平分线上
∴$AD$垂直平分线段$CE$