Question 101
Let $ABCD$ be a rectangle. Let $\omega$ be the semicircle with diameter $[BC]$, such that the point $A$ and the semicircle $\omega$ lie on the same side of the segment $[BC]$. The circle with centre $B$ and radius $AB$ meets the semicircle $\omega$ again at the point $E$. The line $(AE)$ meets the semicircle $\omega$ again at the point $F$. Prove that $AF = BF$.