Monge's theorem states that for any three circles in a plane, none of which is completely inside one of the others, the intersection points of each of the three pairs of external tangent lines are collinear. The article goes on to say that this can be proved easily using Menelaus' theorem.
I understand Menelaus' theorem but don't quite understand how to apply here. Do we consider the triangle $ABC$ formed by joining the centers of the three circles? Then we can consider the three points of intersection $A', B', C'$, each one on the side opposite the respective vertex. It's easy to see that the sign of the product $\frac{AC'}{C'B}\cdot \frac{BA'}{A'C}\cdot \frac{CB'}{B'A}$ is negative. But how to prove that its magnitude is $1$ in order to establish the result?
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$\begingroup$Consider two circles centred at $P,Q$ and their external common tangent intersect at $T$. Then from the diagram above it is quite clear that $\triangle TSQ\sim \triangle TRP$ so $\frac{PT}{QT}=\frac{PR}{QS}=\frac{r_P}{r_Q}$ where $r_P$ and $r_Q$ are the radii of the two circles.
Now onto the problem:
You want to show that the magnitude of $\frac{AC'}{C'B}\cdot \frac{BA'}{A'C}\cdot \frac{CB'}{B'A}$ is 1.
But from the lemma shown above,
the magnitude of $\frac{AC'}{C'B}$ is $\frac{r_A}{r_B}$.
the magnitude of $\frac{BA'}{A'C}$ is $\frac{r_B}{r_C}$.
the magnitude of $\frac{CB'}{B'A}$ is $\frac{r_C}{r_A}$.
Hence, $$\left|\frac{AC'}{C'B}\cdot \frac{BA'}{A'C}\cdot \frac{CB'}{B'A}\right|=\left|\frac{r_A}{r_B}\cdot \frac{r_B}{r_C}\cdot \frac{r_C}{r_A}\right|=1$$
and the conclusion follows.
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