The proof follows at once, by observing the middles of sides of one triangle formed by three (out of the four) sides of the quadrilateral. Take for example triangle BCG. The lines through its middles pass through the three points K, H, I. To prove them collinear it suffices to apply Menelaus theorem to B'C'G' and prove that (*) (HG'/HB')(IB'/IC')(KC'/KG')=1.

Applying the same criterium to triangle BCG and the line DE we have: (AB/AG)(EC/EB)(DG/DC)=1. But the ratios are equal one to one: HG'/HB'=AB/AG, IB'/IC'=EC/EB and KC'/KG'=DG/DC, thus proving (*) (proof taken from Deltheil and Caire).

Triangle XYZ, formed by the diagonals of the c.q. is called

For a property of the Newton line, in the particular case of quadrangles circumscriptible to a circle see Newton2.html .

For another application of Newton's theorem in the case of cyclic quadrilaterals see the file CyclicProjective.html .

IsogonalGeneralized.html

Newton2.html

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