$\angle AFB=2\angle ACB=\angle AOB$ implies $AFOB$ is cyclic. In fact, we will show the $A=60$, $B=75$, and $C=45$, which proves the problem since $\angle AMB=90=\angle AOB$ and $\angle BAM=30=\angle SMK$. Since $\angle BCJ=\angle CBK$, we get $BCKJ$ is an isosceles trapezoid, so $PQ\parallel BC$. Therefore, $Q$ lies on the $A$-altitude. The angle conditions also imply $\angle ABC=5\angle ABE$, so $3B=5C$. This means $C=\frac35B$ so $\angle ABE=\frac15B$. Therefore, $\angle ABQ=\frac25B=\angle ACQ$, so the circumcircle of $BCF$ intersects $AB$ on $CQ$. Therefore, by Radical Axis, the circle through $A$, $C$, and $CQ\cap AB$ passes through $AQ\cap BC$, which is the foot of the $A$-altitude, so $CQ$ must also be an altitude, implying $Q$ is the orthocenter of $ABC$, so $\angle AFB=90$, which means $C=45$.