(1)证明：延长DE至点F,使得EF=BC连接AF
∵∠AED=90°
∴∠AEF=90°
∵AB=AE
∠ABC=∠AEF=90°
BC= EF
∴△ABC≌△AEF
AC=AF
∴∠BAC=∠EAF
∵∠BAE=90° ∠CAD=45°
∴∠BAC+∠DAE=45°
∴∠EAF+∠DAE=45°
∵AC=AF
∠CAD= ∠DAF=45°
AD=AD
∴△ACD≌△ADF
CD=DE=DE+EF=DE+BC
(2)证明 ∵ ∠B=∠E
∠C=∠F
∴△ABC∽△DEF
AB/DE=BC/EF=AB+BC/DE+EF=1
∴△ABC≌△DEF