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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.

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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 $BC$

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  • $\begingroup$ Don't we need the length of AH to calculate the length of AF? $\endgroup$ – MadChickenMan Mar 24 '15 at 18:56
  • $\begingroup$ @MadChickenMan The length of $AH$ is simply $BA-BH$, both of which are known. $\endgroup$ – G-man Mar 24 '15 at 18:59
  • $\begingroup$ Am I missing something, what is the length of BH? $\endgroup$ – MadChickenMan Mar 24 '15 at 19:03
  • $\begingroup$ $BH=\dfrac{r_1+r_2}{2}$ because $HF$ is the perpendicular bisector of $BC$ $\endgroup$ – G-man Mar 24 '15 at 19:04
  • $\begingroup$ Ah yes, well thank you for all your help. By the way, what software is it you used to construct your diagram, I've seen it before, but I'm not sure of the name of it. $\endgroup$ – MadChickenMan Mar 24 '15 at 19:07

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