fuhrmann's theorem

proof
This is obtained by applying ptolemy's theorem to the
convex cyclic quadrilaterals ABDE, BCDF, ADEF, ABEF

|AD||BE| = |AB||DE|+|AE||BD|,
|BD||CF| = |BC||DF|+|BF||CD|,
|AE||DF| = |AD||EF|+|AF||DE|,
|AE||BF| = |AB||EF|+|AF||BE|.

Multiplying the first by |CF|, we get
|AD||BE||CF| = |AB||DE||CF|+|AE||BD||CF|.
By the second, we can replace |BD||CF| by
|BC||DF|+|BF||CD| in the second factor on the right :
|AD||BE||CF| = |AB||DE||CF|+|AE||BC||DF|+|AE||BF||CD|.
Using the third and fourth, we can replace |AE||DF| by
|AD||EF|+|AF||DE|, and |AE||BF| by |AB||EF|+|AF||BE|.
This gives the stated result.

fuhrmann's theorem in hyperbolic geometry
Suppose that ABCDEF is a convex cyclc hexagon.
Then s(AD)s(BE)s(CF) =
s(AB)s(CD)s(EF)+s(AF)s(BC)s(DE)+s(AB)s(DE)s(CF)+
s(BC)s(EF)s(AD)+s(CD)s(AF)s(BE).