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 ÐACB = a (where a is constant) and ÐAEB = ÐAFB = b (where b is constant).
Therefore triangles CAF and CBE are similar. As the angles
in a triangle add up to 180o,
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ÐCAF = ÐCBE = 180 - a- b. |
|
Hence, as angles on a line add up to 180o
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.