Two circles tangent to each other and tangent to a line Two circles with centers $A$ and $B$ are externally tangent at point $C$.
 tangent to the two circles. Given that the radii of the two circles are $2cm$ and $3cm$, respectively, find $\frac{DC}{FC}$

 A: Adjusting Matthew's figure:
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Use Cosine theorem:
$$\frac{FC}{DC}=\frac{\sqrt{3^2+3^2-2\cdot 3^2\cdot \cos \beta}}{\sqrt{2^2+2^2-2\cdot 2^2\cdot \cos (180^\circ-\beta)}}=\sqrt{\frac{18-18\cdot \frac15}{8+8\cdot \frac15}}=\sqrt{\frac32}.$$
Note: $\cos \beta =\frac{BE}{AB}=\frac{FB-FE}{AC+BC}=\frac{FE-AD}{AC+BC}=\frac{3-2}{2+3}=\frac15$.
A: The answer is in fact $\frac{DC}{FC}=\sqrt{\frac{2}{3}}$. To see this, consider the following: 
Since $AD, FB\perp DF$, we have $\alpha+\beta=180^{\circ}$. Now, as $\angle DFC$ is the tangent chord angle wrt the chord $FC$, it holds that $\angle DFC=\frac{\beta}{2}$. Analogously, $\angle CDF=\frac{\alpha}{2}$. Therefore, since $\frac{\alpha}{2}+\frac{\beta}{2}=90^{\circ}$, we have $\angle FCD=90^{\circ}$.
Now, let $P$ be the second point of intersection of the line $FB$ with the circle around $B$. Then $\angle PCF=90^{\circ}$ and because $\Delta BCF$ is isosceles, $\angle CFP=\angle CFB=90^{\circ}-\frac{\beta}{2}=\frac{\alpha}{2}$. Thus, $\Delta PFC\sim\Delta FCD$. 
On the other hand, $\angle CBP=180^{\circ}-\beta=\alpha$ and thus $\Delta PBC$ is isosceles with an apex angle of $\alpha$, as is $\Delta CDA$. Hence, $\Delta PBC\sim\Delta CDA$. Therefore, 
$$\frac{PC}{CD}=\frac{BC}{CA}\Leftrightarrow PC=CD\cdot\frac{BC}{CA}$$
Moreover, from $\Delta PFC\sim\Delta FCD$, we obtain
$$\frac{PC}{FC}=\frac{FC}{CD}\Leftrightarrow \frac{CD\cdot\frac{BC}{CA}}{FC}=\frac{FC}{CD}\Leftrightarrow \frac{BC}{AC}=\bigg(\frac{FC}{CD}\bigg)^2$$
Thus, in this particular case, $\frac{CD}{FC}=\sqrt{\frac{AC}{BC}}=\sqrt{\frac{2}{3}}$.
