∠∠ABC =∠AEP，∠BAC=∠EAP，

∴∠ACB=∠APE，

In ABC, AB=BC,

∴∠ACB=∠BAC，

∴∠epa=∠eap；

(2) Answer: □APCD is rectangular,

∵ Quadrilateral APCD is a parallelogram,

∴AC=2EA,PD=2EP，

∫from( 1)≈EPA =∠EAP，

∴EA=EP, then AC=PD,

∴□APCD is rectangular;

(3) a: EM=EN，

EA = EP，

∴∠EPA=90 -α，

∴∠eam= 180-∠EPA = 180-(90-α)= 90+α，

It can be known from (2) that ∠ CPB = 90, f is the midpoint of BC,

∴FP=FB，

∴∠FPB=∠ABC=α，

∴∠epn=∠epa+∠apn=∠epa+∠fpb=90-α+α= 90°+α，

∴∠EAM=∠EPN，

∠∠AEP rotates clockwise around point E by an appropriate angle to get ∠ ∠MEN,

∴∠AEP=∠MEN，

∴∠AEP-∠AEN=∠MEN-∠AEN，

That is ∠MEA=∠NEP,

∴δeam≌δepn，

∴EM=EN。