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 )
Thanks in advance for any contributions.
Let the radii of the circles be $r_1$ and $r_2$ respectively and let the length $CG$ be $r_2+x$. Since $\triangle CGE\sim\triangle BGD$, $\dfrac{CE}{CG}=\dfrac{BD}{BG}$ or :$$\dfrac{r_2}{r_2+x}=\dfrac{r_1}{r_1+2r_2+x}\implies x=\dfrac{2r_2^2}{r_1-r_2}\implies CG=\dfrac{r_2(r_1+r_2)}{r_1-r_2}$$
Also since $\triangle AHF\sim\triangle CEG$, $\dfrac{AH}{AF}=\dfrac{CE}{CG}$. Now use this to calculate the length of $AF$ and compare it with the length of $BC$