IGO 2016 A2

In acute-angled triangle $ABC$ , altitude of $A$ meets $BC$ at $D$ , and $M$ is midpoint of $AC$ . Suppose that $X$ is a point such that $\measuredangle AXB = \measuredangle DXM =90^\circ$ (assume that $X$ and $C$ lie on opposite sides of the line $BM$ ). Show that $\measuredangle XMB = 2\measuredangle MBC$ .Proposed by Davood Vakili