Congratulations to **Edward, Graveney School, Tooting** for cracking yet another of the geometry problems. Well
done Edward.

As $C$ moves so do the points $E$ and $F$ but the common chord $AB$ to the two circles remains fixed. Angles in the same segment are equal, so $\angle ACB = \alpha$ (where $\alpha$ is constant) and $\angle AEB = \angle\ AFB = \beta$ (where $\beta$ is constant). Therefore triangles $CAF$ and $CBE$ are similar. As the angles in a triangle add up to $180^o$, $$ \angle CAF = \angle CBE = 180 - \alpha - \beta. $$ Hence, as angles on a line add up to $180^o$ $$ \angle\ EAF = \angle EBF = \alpha + \beta. $$ Since equal angles at the circumference of a circle are subtended by equal chords it follows that $EF$ is a chord of constant length in its circle.