[alogo] Newton's theorem

Consider the [complete quadrilateral] formed by four lines, no three of which are concurrent. The four lines are called [sides] of the complete quadrilateral. The six points of intersecttion of pairs of sides are called [vertices] of the c.q. Pairs of vertices not lying on any common side are called [opposite vertices] of the c.q.. The lines through the three pairs of opposite vertices are called [diagonal lines] of the c.q.. In the figure below the four sides of the c.q. are AB, BC, CD, DA. The diagonals (green) of the complete quadrilateral are the segments AC, BD and EG. Newton's theorem claims that their middles H, K and I are collinear.

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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 diagonal 3-line of the c.q.. The line containing points H,I,K is often called Newton line of the quadrangle.
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 .

Remark The meaning of this property is revealed, I think, in the context of triangle geometry and the quadratic transformations associated to a specific point on its plane. This story is discussed in IsogonalGeneralized.html . See in particular section (6) there.

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