$证明：(1)∵AC⊥BC$
$∴∠ACB=90°=∠E$
$在Rt△ABC和Rt△ADE中$
$\begin{cases}{AB=AD}\\{BC=DE}\end{cases}$
$∴Rt△ABC≌Rt△ADE(\mathrm {HL})$
$∴AC=AE$
$(2)延长AF、BC交于点G$
$∵△ABC≌ △ADE$
$∴∠BAC=∠DAE$
$又∠ABC=∠CAD$
$∴∠CAE=∠CAD+∠DAE$
$=∠ABC+∠BAC= 90°=∠ACB$
$∴BG//AE$
$∴∠G= ∠EAG$
$在△AEF 和△GBF 中$
$\begin{cases}{∠AFE=∠GFB}\\{∠EAF=∠BGF}\\{EF=BF}\end{cases}$
$∴△AEF≌△GBF(\mathrm {AAS})$
$∴AE=BG $
$∵AC=AE，∴BG=AC$
$在△ABG 和△DAC 中$
$\begin{cases}{AB=DA}\\{∠ABG=∠DAC}\\{BG=AC}\end{cases}$
$∴△ABG≌ △DAC(\mathrm {SAS})$
$∴∠G = ∠ACD$
$∵∠ACG=∠ACB=90°，即∠ACD+∠GCD=90°$
$∴∠G+∠GCD=90°，∴AF⊥CD$
