Proof of the existence of the nine point circle


Since CS = SH and CP = PA, PS ia parallel to AH,
Since AP = PC and AM = MB, PM ia parallel to AB,
But AE is perpendicular to CB and so PS is perpendicular to PM -- angle MPS = 90deg

Similarly angle MLS = 90deg.

Therefore the circle with diameter MS passes through P, S, L and M, angle ADC = 90deg so D also lies on the circle.

By similar argument we can show that Q, F, E and R also lie on the circle!

Finally sides PL, LM and MP of triangle PLM are each one half the length of the corresponding side of triangle ABC and it follows that the radius of circle PLM is one half that of circle ABC.


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