(1) solution: the quadrangle BEDF is a diamond.

Ef vertically divides BD.

∴BO=DO BD⊥EF

Rectangular ABCD

∴DC∥AB

∴∠CDB=∠ABD

∠∠DOF =∠BOE· Bo =DO

∴△DOF≌BOE

∴DF=BE

∫DF∨BE

Parallelogram bed

∵BD⊥EF

Diamond ·BEDF

(2) Solution: As shown in the figure, in the diamond-shaped EBFD, BD=20, EF= 15.

DO= 10，EO=7.5 ..

De = EB = BF = FD = 12.5 obtained from Pythagorean theorem.

S diamond EBFD= 1/2EF? BD=BE？ AD, (here is bottom × height)

That is,1/2× 20×15 = 25/2× ad.

So we get AD = 12.

According to Pythagorean theorem, AE=3.5, AB = AE+EB = 16.

From 2 (AB+AD) = 2 (16+12) = 56,

So the circumference of the rectangular ABCD is 56.