Let ABCD be a trapezoid, with AB parallel to CD (the vertices are listed in cyclic order). The

diagonals of this trapezoid are perpendicular to one another and intersect at O. The base

angles DAB and CBA are both acute. A point M on the line segment OA is such

that BMD = 90degrees, and a point N on the line segment OB is such that ANC = 90degrees.

Prove that triangles OMN and OBA are similar.

Cheers