RU-SPB 2019 Problem 104

Given a convex quadrilateral $ABCD$ . The medians of the triangle $ABC$ intersect at point $M$ , and the medians of the triangle $ACD$ at point $N$ . The circle, circumscibed around the triangle $ACM$ , intersects the segment $BD$ at the point $K$ lying inside the triangle $AMB$ . It is known that $\angle MAN = \angle ANC = 90^o$ . Prove that $\angle AKD = \angle MKC$ .