Let $ABC$ be an acute triangle with $AB<AC$. Point $M{}$ from the side $(BC)$ is the foot of the bisector from the vertex $A{}$. The perpendicular bisector of the segment $[AM]$ intersects the side $(AC)$ in $E{}$, the side $(AB)$ in $D$ and the line $(BC)$ in $F{}$. Prove that $\frac{DB}{CE}=\frac{FB}{FC}=\left(\frac{AB}{AC}\right)^2$.