@e2718pi3142 As ΔABO=ΔDCO by two sides and the angle between them, angles AOD and BOC are equal. As COBF is cyclic, angles AFD and COB are also equal, which makes AFOB cyclic. Then, CB is the O-Simson line of ΔAFD and OE is the altitude of isosceles ΔAOD with AO=OD. Thus, AE=ED.
@e2718pi3142 arc AB = arc CD (same angles at circumference) so AB = DC
r = semiperimeter - hypot = (8+4+4*sqroot3)/2 - 8
= 2*sqroot3 -2
diameter is 4*sqroot3 -6