Because: AE is a crease, and point D falls on point F.

So: AE divides DF vertically.

So: angle ADF=∠AFD, ∠EDF=∠EFD.

So: ∠ AFE = ∠ Ade = 90.

So: ∠ AFB+∠ EFC = 90.

And: < FEC+< EFC = 90.

So: ∠AFB =∠ wild card

So: RT△AFB∽RT△FEC

According to RT△AFB∽RT△FEC, AB/FC=BF/EC.

Namely: AB/BF=FC/EC=4/3.

Let EC=3x, then CF=4x, EF=DE=5x.

So: CD=8x=AB

So: BF=(3/4)AB=6x.

So: AD=BC= 10x.

In RT△AED, from Pythagorean theorem: 25*5= 100x+25x.

The solution is x= 1.

So: AB=DC=8, AE=BC= 10.

So: the circumference of rectangular ABCD is 8+8+ 10+ 10=36.