In a convex quadrangle $ ABCD $, the rays $ DA $ and $ CB $ intersect at point $ Q $, and the rays $ BA $ and $ CD $ at the point $ P $. It turned out that $ \angle AQB = \angle APD $. The bisectors of the angles $ \angle AQB $ and $ \angle APD $ intersect the sides quadrangle at points $ X $, $ Y $ and $ Z $, $ T $ respectively. Circumscribed circles of triangles $ ZQT $ and $ XPY $ intersect at $ K $ inside quadrangle. Prove that $ K $ lies on the diagonal $ AC $.