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.