Let the points $A,B,C$ lie on a line in this order. $AB$ is the diameter of semicircle $\omega_1$, $AC$ is the diameter of semicircle $\omega_2$. Assume both $\omega_1$ and $\omega_2$ lie on the same side of $AC$. $D$ is a point on $\omega_2$ such that $BD\perp AC$. A circle centered at $B$ with radius $BD$ intersects $\omega_1$ at $E$. $F$ is on $AC$ such that $EF\perp AC$. Prove that $BC=BF$.