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Two opposite angles of a quadrilateral equal

Source: 5th QEDMO problem 2, proposed by me, discussed elsewhere

November 10, 2007
geometrytrapezoidcircumcirclegeometry proposed

Problem Statement

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