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.
Best Answer
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.