Wow,the point $F$ has a tone of properties ! First let's add some points: Let $X$ such that $ACBX$ is parrallelogram,let ${S}=AB'\cap BC$ and take the two points $C'$ and $C_1$ that make the two triangles congruent such that $A,C',B'$ are in counterclockwise order and the other in clockwise order.We have to prove two things that finish the problem : 1)$\angle B'C'C=\angle CAF$ 2) $\angle B'CC_1=90^{\circ}$ Note that $AB'=BC=AX$ so $AB'X$ is isosceles for 1) we need the following lemma: Lemma(better characterization for $F$): we have a)$FB'=FC$ b)$\angle FCB=90^{\circ}$ c) $F\in (B'SC)\cap (ABS)$ d)$FA=FB$ Proof: let's redefine $F$ as the intersection between the perpendicular from $C$ to $BC$ and from $B'$ to $B'A$. Observe that $\angle FCB'=90-\angle B'CB=90-\angle AXB'=90-\angle AB'X=\angle FB'C$ so $FB'=FC$.Also since $\angle FCS=\angle FB'S=90$ we clearly have $FB'CS$ cyclic Now observe again that $\triangle FCB \equiv \triangle FB'A$ so $FA=FB$.We also have from this that $FAB \sim FB'C$ so $\angle FAB=\angle FB'C=\angle FSC$ implies $FABS$ cyclic. Also from the similarity we have $\angle FCM=\angle FCB'=\angle FBA=\angle FBM$ so $F\in (BMC)$ implying $F$ is the original point.The lemma is proved. $\blacksquare$ Coming back to our problem let's prove 1) first. The idea is that $CC'B' \sim CAF$.To prove that just note $\frac{CA}{AC'}=\frac{CF}{FB'}=1$ and $\angle CAC'=\angle CAB'+\angle C'AB'=\angle ACB+\angle CAB'=\angle ASB=\angle B'FC$ so $CC'A \sim CB'F$ and by spiral we get $CB'C' \sim CFA$ and we are done. For 2) angle chase again: $\angle B'CC_1=\angle B'CA+\angle ACC_1=\angle MCA+\angle 90-\angle CC'C_1=\angle MCA+90-\angle CC'B'-\angle B'C'C_1=\angle MCA+90-\angle CAF-(90-\angle C'B'A)=\angle MCA+\angle CBA-\angle CAF=\angle MCA+\angle CBA-(\angle BAF-\angle BAC)=\angle MCA+\angle CBA-\angle BAF+\angle BAC=180-\angle ACB+\angle MCA-(90-\frac{\angle ASB}{2})=90-\angle MCB+\frac{\angle ASB}{2}=90$ done. The problem is solved. $\blacksquare$

Throughout this solution we will redefine the following points and show that they satisfy the conditions of the question. (Notice that the last definition proves the desired equality): $F$ is the antipode of $B$ with respect to the circumcircle of triangle $\Delta BMC$ $B'$ is the point along $CM$ such that $FC=FB'$ $C'$ is the point such that triangles $\Delta CFA$ and $\Delta CB'C'$ are similar Claim: $\angle FB'A=90^{\circ}$ Since $\angle BMF=90^{\circ}$ we get $FA=FB$. As $\angle ABF=\angle B'CF$ isosceles triangles $\Delta ABF$ and $\Delta B'FC$ are similar. By SST triangles $\Delta AB'F$ and $\Delta BCF$ are also similar, proving the claim. Claim: Triangles $\Delta ABC$ and $\Delta C'B'A$ are congruent Notice as $AF=BF$ we must have $AB'=BC$. Also by SST triangles $\Delta ABF$, $\Delta B'FC$, and $AC'C$ are similar. Thus $AC=AC'$ and it is sufficient to show that $\angle AC'B=\angle BAC$. This is just an angle chase: $$\angle AC'B'=\angle AC'C-\angle B'C'C=\angle BAF-\angle CAF=\angle BAC$$

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Note that by law of sine, $AMB’$ and $BMC$ have the same circumradii. Assume that the circumcircle of $AMB’$ meet the circumcircle of $BMC$ again at $F’$. Since $AM=MB$, we have $\angle AF’M=\angle MF’B$, implying $\angle AB’F’=\angle AMF’=90^{\circ}$, i.e. $F’=F$. Moreover, we also obtain that $\triangle AB’F \simeq \triangle BCF$ and $\triangle AFB\sim \triangle B’FC$. Hence, $$\dfrac{C’B’}{AF}=\dfrac{AB}{AF}=\dfrac{AB}{FB}=\dfrac{B’C}{FC}.$$ Moreover, by angles chasing, we also obtain that $\angle C’B’C=\angle AFC$. Hence, $\triangle C’B’C\sim \triangle AFC$, so we are done.

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