Belorusija 2012

Point $M$ is marked inside the convex quadrilateral $ABCD$ so that the ratio of the areas of the triangles $AMC$ and $BMD$ is equal to the ratio of the tangents of the angles $AMC$ and $BMD$, i.e. $S(AMC):S(BMD)=\mathrm{tg}\angle AMC:\mathrm{tg}\angle BMD$. Prove that $A{M}^2+M{C}^2+B{D}^2=A{C}^2+B{M}^2+M{D}^2$ if $M$ does not belong to any of the diagonals of the quadrilateral.