[alogo] 1. Quadrilateral properties

Let p = ABCD be a complete quadrilateral, {E, F, G} its diagonal points and q = HIKL the quadrangle having vertices the intersection points of the diagonals with the sides of ABCD.
[1] The diagonal GH intersects the parallel DN from D to the side HI of q at the middle M of ND.
[2] Side HJ of q intersects segment MN at its middle O.


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[1] MN/MD = 1, since by Menelaus theorem applied to triangle AND with secant line HMG:
(HN/HA)*(MD/MN)*(GA/GD)=1. But HN/HA=ID/IA=GD/GA.
[2] Also ON/OM = 1, since the bundle H(J,I,E,F) is harmonic. Thus the parallel NM to line HI of the bundle is divided in two equal parts by the other two.

[alogo] 2. Newton line property

Let p = ABCD be a complete quadrilateral, {E, F, G} its diagonal points and q = HIKL the quadrangle having vertices the intersection points of the diagonals with the sides of ABCD.
Consider the middles {P, S, R} of the diagonals of the quadrilateral, which are collinear points contained in the Newton line of the quadrilateral.
[1] The medians {AA1, BB1, CC1, DD1} of the triangles respectively {AHI, BJH, CKJ, DIK} meet at a point S on the Newton line.
[2] S is the harmonic conjugate of one diagonal middle (R) with respect to the two others (P, Q).


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[1] Start with the intersection point T of diagonal EF with line AR. Draw from T line TV parallel to side AB intersecting side CD at U. Since the bundle F(V,T,U,A) is harmonic and TV is parallel to ray FA of it U is the middle of TV.
[2] Since D(A,W,T,R) is a harmonic bundle and R is the middle of FG, its ray DT is parallel to FG.
[3] It follows from [1,2] that DTFV is a parallelogram. Thus, U is the middle of DF, hence the initial parallel TV to line AB passes through the middles of segments having one end-point at D and the other on line AB. Among them it passes through the middles of {DA, DN, DB} the last being P: the middle of the diagonal DB.
[4] Extend the median AA1 of triangle AHI to intersect the Newton line at S. Bundle A(P,Q,S,R) is harmonic. In fact, it has the same traces on line TV with the harmonic bundle E(P,A,M,T). Thus, S is the harmonic conjugate of R with respect to {P,Q}.


See Also

Newton.html
Newton2.html

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