and any two other chords DE and FG going through C.

FD and FE intersect AB at M and N.

C is also the midpoint of MN

An elementary proof :

The inscribed angles <)DFG = <)DEG and <)FDE = <)FGE

Triangles FCD and ECG are then similar and FD/FC = EG/EC

Draw the perpendicular OH _|_ DF and OJ _|_ EG

H is the midpoint of chord DF :

FD = 2.FH, and similarily EG = 2.EJ, that is 2.FH/FC = 2.EJ/EC

or FH/FC = EJ/EC and as <)HFC = <)JEC,

triangles FCH and ECJ are similar, hence <)FHC = <)EJC

Hence inscribed angles <)MHC = <)MOC, and also <)NOC = <)NJC hence <)MOC = <)NOC

The right triangles MOC and NOC are then equal and CM = CN

QED.