[alogo] 1. Applying Menelaus to a paralleogram problem

Consider a parallelogram EFGH and two points I, J on two opposite sides. The intersection-points K, L of the two triangles HJG and EIF
define a line e = [KL] passing through the center of the parallelogram ([Anthony, p. 77], [Papelier, Rapport anharmonique p. 62]).

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Apply Menelaus theorem (see Menelaus.html ) two times for the line e and the two triangles :
(1)  EIF  => (MF/ME)(KE/KI)(LI/LF) = 1,
(2)  HJG => (NH/NG)(LG/LJ)(KJ/KH) = 1.
Divide the sides of the equations and note that KE/KI = KJ/KH, LI/LF = LG/LJ => MF/ME = NH/NG.
Later implies that MF = NH, hence line e passes through the center  X of the parallelogram.
Remark Essentially the exercise is a special case of the famous theorem of Pappus on the collinearity of three intersection points (see
PappusLines.html ). The general case of this theorem reduces to the present particular case by using a special projectivity that sends two points
to infinity.

[alogo] 2. Intersecting diagonals

Consider a parallelogram ABCD and an arbitrary point E. Draw from E parallels to the sides to build parallelograms AHEI and FCGE.
The diagonals {HI,FG,BD} of these parallelograms and the original one intersect at a point J.

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Consider J as the intersection point of the two lines {BD, FG} and apply Menelaus theorem to triangle BCD and its secant FG:
(DG/GC)*(CF/FB)*(BJ/JD)=-1.Notice that DG/GC=AH/HB, CF/FB=DI/IA, hence(DI/IA)*(AH/HB)*(BJ/JD)=-1. This, by applying the
inverse Menelaus theorem to triangle ABD, implies that HI passes through J too.

[alogo] 3. Intersecting diagonals, the general case

The previous property is a special case of a property of general quadrilaterals. In fact, consider an arbitrary quadrangle q=ABCD and
an arbitrary point E. Join point E to two vertices {K,L} of q and consider the quadrilateral q'=FGIH defined by the intersections of
lines {EK,EL} with the sides of the original quadrilateral. The pairs of opposite sides of q' intersect on the diagonals of q.

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The figure shows only one pair (HI,FG) of opposite sides of q' (the other being (HF,GI)) intersecting at J. The property follows by
transforming the quadrilateral to a parallelogram via a projectivity and applying the results of (2).

See Also

Menelaus.html
PappusLines.html

Bibliography

[Anthony] O. W. Anthony Problem 36 American Mathematical Monthly v. 2(1895)
[Papelier] Papelier, G. Exercices de Geometrie Moderne. Paris, Vuibert, 1927

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