$ (1)$解：由题意得$\triangle ABC≌\triangle DEC$
∴$BC = EC，$$∠ACB=∠DCE$
∵$BC = 2，$$∠ACB = 90°，$∴$EC = 2，$$∠DCE = 90°$
∴$∠ACB+∠DCE = 180°，$即$A，$$C，$$E$三点在同一条直线上
$ $又$AE = 6，$∴$AC=AE - EC=6 - 2 = 4$
$ (2)$证明：延长$AB$交$DE$于点$F$
∵$∠ACB = 90°，$∴$∠ABC+∠BAC = 90°$
∵$\triangle ABC≌\triangle DEC，$∴$∠EDC=∠BAC$
∵$∠ABC=∠DBF$
∴$∠DBF+∠EDC = 90°，$即$∠AFE = 90°$
∴$AF\perp DE，$即$AB\perp DE$
