Analysis: According to the trapezoid midline theorem, the trapezoid midline is equal to half of the sum of the upper and lower bottoms of the trapezoid. If the center line is EF, then EF= 1/2(AB+CD), that is, 2EF=AB+CD is inscribed with circle O at m, n, e and f, and it is known that DN = DM, cm = cf and BF = be.

Solution: Let the center line of this trapezoid be EF.

Then EF= 1/2(AB+CD).

That is 2EF=AB+CD.

Because EF=8

So AB+CD= 16

And because the circle O is inscribed with the isosceles trapezoid ABCD in M, N, E and F.

Connect mo, no and do respectively.

Then MO = no

∠OMD=∠0ND=90 degrees

Because DO=DO

So △ DMO△ DNO

So DN=DM

Similarly CM=CF, BF=BE, AE=AN.

The circumference of the trapezoid C=AB+BC+CD+DA.

=2(AE+BE+DM+CM)

=2(AB+CD)=2× 16=32