Congratulations to Edward Wallace, 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 ∠ACB = α (where α is constant) and ∠AEB = ∠AFB = β (where β is constant). Therefore triangles CAF and CBE are similar. As the angles in a triangle add up to 180^o,

∠CAF = ∠CBE = 180 − α− β.

Hence, as angles on a line add up to 180^o

∠EAF = ∠EBF = α+ β.

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.