$AB$ 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.