# 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.

• In your figure $O$ lies on line $O_1O_2$. Is this an additional hypothesis? – Matteo Mar 21 at 22:07
• Yes! I forgot that – user2471 Mar 21 at 22:21
• that explains the counter example below... – Matteo Mar 21 at 22:22
• @Matteo my Apologies! – user2471 Mar 21 at 22:26
• 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... – CiaPan Mar 21 at 22:52

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.

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...

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

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.

• OP requires not to use trigonometry. – Matteo Mar 21 at 22:25
• @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 $\frac{IB}{2R_1}=\frac{IC}{2R_2}$. – user Mar 21 at 22:47

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.