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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.

enter image description here

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  • $\begingroup$ In your figure $O$ lies on line $O_1O_2$. Is this an additional hypothesis? $\endgroup$ – Matteo Mar 21 at 22:07
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    $\begingroup$ Yes! I forgot that $\endgroup$ – user2471 Mar 21 at 22:21
  • $\begingroup$ that explains the counter example below... $\endgroup$ – Matteo Mar 21 at 22:22
  • $\begingroup$ @Matteo my Apologies! $\endgroup$ – user2471 Mar 21 at 22:26
  • $\begingroup$ If you name the endpoints of the green diameter say M and N, then $\angle MBI$ and $\angle NCI$ are both right, hence right triangles $\triangle MBI$ and $\triangle NCI$ are similar (with the scale equal to the ratio of radii of small circles); then $I$ divides $BC$ in the same proportion as it divides the green diameter $MN$. Alas, I can't see yet how it can help... $\endgroup$ – CiaPan Mar 21 at 22:52
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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.

enter image description here

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.

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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...

enter image description here

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  • $\begingroup$ My apologies! I forgot one crucial assumption: colinearity of the circle centers! $\endgroup$ – user2471 Mar 21 at 22:28
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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.

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  • $\begingroup$ OP requires not to use trigonometry. $\endgroup$ – Matteo Mar 21 at 22:25
  • $\begingroup$ @Matteo I have noticed it. I however do not know if it refers to the usage of cosine law (or Pythagorean theorem). No other "trigonometric" properties are used. In fact $\cos\alpha$ can be considered here as an abbreviation for $\frac{IB}{2R_1}=\frac{IC}{2R_2}$. $\endgroup$ – user Mar 21 at 22:47
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Here is a possibile path.

enter image description here

  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.

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