Butterfly Theorem

Given circle with center O, a chord AB with midpoint C
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

OCMH cocyclic, on circle with diameter OM (angles at C and H are right angles)
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.

 

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