[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.

[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.

Consider the middles {P, S, R} of the diagonals of the quadrilateral, which are collinear points contained in the

[1] The medians {AA

[2] S is the harmonic conjugate of one diagonal middle (R) with respect to the two others (P, Q).

[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 AA

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