## Equi-areal trapezium division

From the middle F, of a non-parallel side of trapezium t = ABCD, draw a line FE, dividing the trapezium in two quadrangles with equal areas.

Take DG = AB and parallel to it. Define parallelogram p = BCIH, with center of symmetry J, the middle of AD. Ç GH, defines E, as its intersection point with AD. It is (AE/ED) = (AH/AB) = (DC/AB) and triangles DCE and EAB have equal areas.
Show also that KD // EC, KL // EB and CM = BL.
Considering the trapezium as the "half" of the parallelogram p, the problem is related to the division problem for parallelograms: Divide a parallelogram in four equal (in area) parts, by intersecting it with two lines, passing through a common point. Look at the file ParaDivision.html , for a discussion of the subject.
According to that discussion, the line FK, found here, is the conjugate direction of BC, with respect to the minimal circumellipse of the parallelogram, resulting by taking the symmetric of t, w.r. to F.

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