Students will learn what Fuhrmann's theorem is, and why it's true.

Theorem: Suppose that ABCDEF is a convex cyclic hexagon. Then |AD||BE||CF| = |AB||CD||EF| + |AF||BC||DE| + |AB||DE||CF| + |BC||EF||AD| + |CD||AF||BE| 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.