Point $F$ lies on the straight line $AC$, with which a circle centered by $F$ can be made through both $B$ and $C$. $M$ is the midpoint of the arc $BC$. The straight line $CM$ intersects the straight line $AB$ at H.
How to express the ratio of the areas of the $\triangle{CBM}$ to $\triangle{CBH}$ in the following three variables.
a) $\alpha$ : the angle $BAC$.
b) $m$ : the length of the segment $AB$.
c) $t$ : the tangent of $\angle{ACB}$.
I've tried the rules of cosines but it ain't working.
This is not an answer. Just a figure for clarification
Here is another case of figure with $F$ outside line segment $[A,C]$ ; do you accept it ?