Denote $\vec{AE} = x$ and $\vec{BD} = y$. Obviously, we have $4x + y = 180^\circ$.
From $\angle APM + \angle MAP = 2x + \frac{y}{2} = 90^\circ$ we deduce that $AM \perp MP$. Since $FO$ is the perpendicular bisector of $BC$, we have $FB = FC$, thus the triangle $FBC$ is isosceles, with the apex $F$. From $\angle BCJ = \angle CBK = \frac{x}{2}$, we deduce that the triangles $BCQ$ and $CBP$ are congruent, therefore $PQ \parallel BC$.

Since $FBC$ is an isosceles triangle, with the base $[BC]$, the line $FO$ is the perpendicular bisector of $PQ$. In the right angled triangle $AQP$, the perpendicular bisector of $PQ$ and the hypotenuse intersect at $F$, therefore $F$ is the midpoint of $AP$.
Denote by $L$ the intersection point of the half-line ($BF$ with the circle $C(O)$).
Because $\angle BFD = \angle CFD$, we have $\vec{BD} + \vec{EL} = \vec{CD} + \vec{AE}$. From $\vec{BD} = \vec{CD} = \angle A$ we obtain $\vec{AE} = \vec{EL} = x$, therefore $\angle ABL = \angle LBK = x$. Since $F$ is the midpoint of $AP$, the line $BF$ is at the same time median and angle bisector of the triangle $ABP$, thus $BA = BP$.
From $\angle ABP = \angle APB = 2x$ and $AM \perp MP$ we deduce that $AB = AP$, therefore the triangle $ABP$ is equilateral. Thus, we obtain $\angle A = 60^\circ$, $\angle B = 75^\circ$ and $\angle C = 45^\circ$.
Moreover, $\angle AOB = \angle AFB = \angle AMB = 90^\circ$, so the points $A, F, O, M, B$ lie on the circle $\omega$ of diameter $AB$. Therefore $\angle BAM = \angle SMK = 30^\circ$, which means that the line $MS$ is tangent to the circle $\omega$.