Geometry · Convex quadrilaterals · Perpendicular bisectors · Congruent triangles · Angle chasing
Let \(ABCD\) be a convex quadrilateral which is not a trapezoid. The perpendicular bisectors of the sides \(AD\) and \(BC\) meet at a point \(P\), and the perpendicular bisectors of the sides \(AB\) and \(CD\) meet at a point \(Q\). Suppose that \(P\) and \(Q\) both lie in the interior of \(ABCD\) and that
\[ \angle APD = \angle BPC . \]
Prove that \(\angle AQB = \angle CQD\).
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Serbian Regional Competition 2012, high school grade I, category A, problem 4. Organized by the Mathematical Society of Serbia (DMS). Source