1) The non-parallel sides extended cut at point G, through which passes line EF, joining the middles of the parallel sides. 2) On EF lies also the intersection point H of the diagonals of the trapezium. 3) Points {E,F,H,G} form a harmonic tetrad (or division see Harmonic.html ). Last property holds also for general quadrangles (see Quadrangle_0.html ).
1) Draw the medial line GE of the triangle GDC and show that it is also medial to triangle GAB. 2) Use Ceva's theorem (see Ceva.html ). 3) HE/HF = DE/FB, GE/GF = DE/AF etc.
In the file ParallelMedians.html we discuss an application of the previous properties to triangles.
1) The diagonals d1=AC, d2 = BD and and sides as shown in the figure satisfy [Rouche, p. 329]: d12 + d22 = c2 + d2 +2ab. 2) It is also [Rouche, p. 330]: (d22 - d12)/(d2 - c2) = (a+b)/(b-a).
To prove (1) write the four following equations combine them to eliminate cosines and divide the result by (a+b):
To show the second property do a similar calculation using triangles AHD and BHC, setting AH = kd1, BH = kd2, HC = k'd1, HD = k'd2, with k/k' = b/a (see section-1).