# BMO1 2009/10 Question 4 Geometry Problem

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 $$BC = 2AF$$.

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 $BC$
• @MadChickenMan The length of $AH$ is simply $BA-BH$, both of which are known. – G-man Mar 24 '15 at 18:59
• $BH=\dfrac{r_1+r_2}{2}$ because $HF$ is the perpendicular bisector of $BC$ – G-man Mar 24 '15 at 19:04