Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

$AB$ is the diameter of $\omega_1$, so its center lies on $AB$.

$BD$ is tangent to $\omega_1$, implies $BD \perp AB$ or $BD\perp FC$.\newline

We want to prove that $BC=BF$, or $B$ is the midpoint of $FC$.

So, we only need to show that $\angle BDF=\angle BDC$.

$\triangle AEB\sim \triangle EFB$, because $\angle EFB=\frac{\pi}{2}$ and $\angle AEB=\frac{\pi}{2}$ since $AB$ is the diameter of $\omega_1$. Also, $\angle FBE=\angle EBA$.

In, $\triangle BAD$ and $\triangle BDF$,

$\angle DBA=\angle FBD$

    $$\frac{BE}{BA}=\frac{BF}{BE}\space\text{[Since, $\triangle AEB\sim \triangle EFB$]}$$

    $$\implies\frac{BD}{BA}=\frac{BF}{BD}\space\text{[$BD=BE$ as they are the radii of the same circle]}$$

    $$\implies \triangle BAD\sim\triangle BDF$$

So, $\angle BDF=\angle BAD=\frac{\pi}{2}-BDA=\angle BDC$ or  $BC=BF$.

Hint:

$\bullet$ Try to show $\angle BDF=\angle BDC$

$\bullet$ Use similarity of triangles.