Let ABCD be a (not self-intersecting) quadrilateral satisfying \measuredangle DAB \equal{} \measuredangle BCD\neq 90^{\circ}. Let X and Y be the orthogonal projections of the point D on the lines AB and BC, and let Z and W be the orthogonal projections of the point B on the lines CD and DA.
Establish the following facts:
a) The quadrilateral XYZW is an isosceles trapezoid such that XY∥ZW.
b) Let M be the midpoint of the segment AC. Then, the lines XZ and YW pass through the point M.
c) Let N be the midpoint of the segment BD, and let X′, Y′, Z′, W′ be the midpoints of the segments AB, BC, CD, DA. Then, the point M lies on the circumcircles of the triangles W′X′N and Y′Z′N.
[hide="Notice"]Notice. This problem has been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=172417 . geometrytrapezoidcircumcirclegeometry proposed