Two circles, of different radius, with centres at $B$ and $C$, touch externally at $A$. A common tangent, not through $A$, touches the first circle at $D$ and the second at $E$. The line through $A$ which is perpendicular to $DE$ and the perpendicular bisector of $BC$ meet at $F$. Prove that $BC = 2AF$.
(source) (British Mathematical Olympiad )
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