Congruent triangles in 3 tangent circle configuration

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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...

Answer 4

Here is a possibile path.

1. $\angle O_1BI \cong\angle O_2CI$ (can you tell why?). Therefore $O_1B\parallel O_2C$.
2. Consequently $\angle BO_1O \cong \angle CO_2O$.
3. $\triangle O_1BO \cong \triangle O_2CO$ (can you tell why?).
4. 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.