Congruent triangles in 3 tangent circle configuration Two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ of centers $O_1$ and $O_2$ are externally tangent at $I$ and internally tangent to a third circle $\mathcal{C}$ of center $O$ that is colinear with $O_1$ and $O_2$ as depicted below. 
A line going through $I$ intersects the three circles at the points $A, B, C, D$ (see figure below). 
How to prove that $AB=CD$ without using trigonometry? 
I tried to show that the triangles $\Delta ABO$ and $\Delta DCO$ are congruent, but I was unable to get the needed angle equalities - $OA=OD$ and $\angle A=\angle D$ are obvious.

 A: After OP added the assumption of collinearity of all three circle centers here is the answer.
Let's draw lines from points M, N, O perpendicular to the chord AD.

Triangles MBI, OSI and NCI are all right and similar. Then S is the midpoint of AD and the lengths between B, I, S and C keep the same proportion as the lengths between M, I, O and N, respectively (they are actually a parallel projection between the lines MN and BC). Hence S is not only a midpoint of AD, but also a midpoint of BC. As a result 
$$AB = AS - BS = \frac 12 AD - \frac 12 BC = DS - CS = DC$$
Q.E.D.
A: I will assume that points $O,O_1,O_2$ are collinear in accordance with your drawing. Otherwise the claim is in general not valid.
Let the points of intersecttion of circle $\mathcal{C}$ with circles $\mathcal{C}_1$ and $\mathcal{C}_2$ be $I_1$ and $I_2$, respectively. Let $\angle O_1IA=\angle O_2IB$ be $\alpha$. Let $R_1$ and $R_2$ be the radii of $\mathcal{C}_1$ and $\mathcal{C}_2$, respectively.
We have
$$
IO=R_2-R_1,\quad IB=2R_1\cos\alpha,\quad IC=2R_2\cos\alpha.
$$
The last two equalities follow from considering the right triangles $IBI_1$ and $ICI_2$.
By the law of cosines one then obtains:
$$\begin{align}
OB^2&=(R_2-R_1)^2+(2R_1\cos\alpha)^2+2(R_2-R_1)(2R_1\cos\alpha)\cos\alpha
=IO^2+IB\cdot IC,\\
OC^2&=(R_2-R_1)^2+(2R_2\cos\alpha)^2-2(R_2-R_1)(2R_2\cos\alpha)\cos\alpha
=IO^2+IB\cdot IC.
\end{align}
$$ 
Thus, $OB=OC$. The rest is simple.
A: Something is wrong. I'm afraid that without some additional assumptions regarding the 'line through I' (or maybe the circles' configuration) your claim is unprovable...

A: Here is a possibile path. 



*

*$\angle O_1BI \cong\angle O_2CI$ (can you tell why?). Therefore $O_1B\parallel O_2C$.

*Consequently $\angle BO_1O \cong \angle CO_2O$.

*$\triangle O_1BO \cong \triangle O_2CO$ (can you tell why?).

*In particular, $OB\cong OC$. Thus $\triangle OBC$ is isosceles.


And the thesis follows from what you already noticed, since now you can demonstrate that $\triangle ABO \cong \triangle CDO$, as you correctly wished to show.
