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Question: Let $o_1$, $o_2$, and $o_3$ be circles with disjoint interiors with centres $O_1$, $O_2$, and $O_3$, respectively. Among the lines tangent to both of the circles $o_2$ and $o_3$ there are two lines that separate $o_2$ from $o_3$ (meaning that $o_2$ and $o_3$ lie on distinct sides of the line). Let $A_1$ be the intersection point of these two lines. Define analogously $A_2$ and $A_3$ using the other pairs of circles. Prove that the lines $A_1O_1$, $A_2O_2$, and $A_3O_3$ intersect at a common point.

This is what I have so far:

The centers of the circles form a triangle. A2 lies on the segment(or side of triangle) o1o3 by the strongest theorem of geometry(o1 lies on the bisector of the angle at A2 and o3 lies on the bisector of the opposite angle at A2. Therefor they all lie on the same line). Similarly, A1 lies on the segment o3o2 and A3 lies on the segment o1o2.

Thus, A1, A2, A3 lie on distinct sides of the triangle o1o2o3. To prove that the lines from each vertex to the opposite point intersect we can use Ceva's theorem.

This is where I am stuck. I am not sure how to show that |o2A3||o1A2||o3A1|= |o1A3||o3A2||o2A1|

I think the strongest theorem of geometry can be used to obtain some equalities with respect to the sides but I don't know where that would help. It feels as though I am missing a small step. Any help is much appreciated.

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    $\begingroup$ $O_3A_1/O_2A_1=r_3/r_2$ where $r$ are the radii. $\endgroup$ – Michal Adamaszek Dec 10 '18 at 9:41
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Let $r_1$, $r_2$, and $r_3$ be the radii of $o_1$, $o_2$, and $o_3$, respectively. Let $i\in\{1,2,3\}$. The indices are considered modulo $3$. Let $S_{i-1}S_{i+1}$ be an internal common tangent of $o_{i-1}$ and $o_{i+1}$ with $S_{i-1}\in o_{i-1}$ and $S_{i+1}\in o_{i+1}$. Prove that $O_{i-1}S_{i-1}T_i$ and $O_{i+1}S_{i+1}T_i$ are similar triangles. This shows that $$\frac{O_{i-1}A_i}{A_iO_{i+1}}=\frac{r_{i-1}}{r_{i+1}}\,.$$

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