# Prove that two segments are congruent in the arbelos

## Background Info + Problem

I teach HS Geometry to middle school age students. I generally like to try to solve problems instead of looking up the answer, but this week a student emailed me a problem that I don't think I'll be able to solve on my own. So I thought I'd ask it here since I haven't been able to find a solution online. The problem is to prove that $\overline{AC} \cong \overline{EF}$. The diagram is known as the arbelos. Information can be found at wolfram|alpha, wikipedia. The green circle in the construction is often referred to as one of the two "twin-circles" or "Archimedes circles." ## Here's a hint

I have researched this, and found the problem presented in a few places. I think that a distorted version of this problem was the inspiration for a study of the arbelos by H.P. Boas. Who, in the end of his work (pdf) poses the problem as an exercise to the reader. He offers the hint that

It suffices to show that $\overline{AF} \perp \overline{EF}$, for then $\triangle AFG \sim \triangle ECG$. Invert in a circle centered at A that cuts the right-hand twin circle orthogonally, and prove that the point F is left fixed.

One sub-question to the main point, is a more general geometric question: Is there a good way to break this proof down into simpler parts? I could keep track of the various ratios throughout the 20+ steps of this construction, but that would be difficult to write and probably more difficult for someone else to follow.

## Construction

If it is helpful, I will present the steps of the construction.

1. Given $\overline{AB}$ with midpoint $C$
2. Construct $\odot C$ with radius $\overline{CA}$
3. Let point $D$ lie on $\overline{AB}$
4. Points $E$ and $F$ are the midpoints of segments $AD$ and $DB$
5. Construct circles $E$ with radius $\overline{EA}$ and $F$ with radius $\overline{FA}$. This is the basic arbelos.

To construct the tangent or "twin-circles"

1. Construct perpendicular lines to segment $AB$ through $E$ and $F$. Let point $G$ be the intersection point of the perpendicular and circle $E$ and point $H$ be the intersection point of the perpendicular and circle $F$.
2. Let $I$ be the intersection of $FG$ and $EH$ 1. Construct circle $D$ with radius $\overline{DI}$
2. Let J and K be the points where circle $D$ intersects segment $AB$
3. Construct perpendiculars to segment $AB$ through $J$ and $K$
4. At the intersection of the perpendicular from K and the circle with center F and radius $\overline{FJ}$, construct circle $L$ with radius equal to $DK$
5. At the intersection of $\overrightarrow{FL}$ and the circle of center $F$ and radius $\overline{FD}$ find point $M$, at the intersection of $\overrightarrow{FL}$ and the perpendicular from D, find point $N$ In the construction I outlined above, I want to prove that $\overline{AD} \cong \overline{MN}$.

# How would you prove $\overline{AD} \cong \overline{MN}$?

• The "special case" of Descartes' Theorem for Kissing Circles lets you compute the radius of the green circle. From here, the problem reduces to algebra. Boas seems to have something else in mind, however; probably something easier. I'm still investigating. :) – Blue Nov 28 '14 at 8:07
• @Blue The radius of the green circle is equal to the radius of circle D with radius $\overline{DK}$. At least to me the Algebra seems like it must get very difficult, but I might have to dig in and see how far I can get. – Nick H Nov 29 '14 at 5:49
• I'm not seeing Boas' intended soln, but here's an approach: Let $X$ be the pt of tangency of the twin circle w/ $CE$, and $Y$ the pt of tangency w/the big semicircle. Introduce $AY$ & $BY$, which form a right angle at $Y$. Convince yourself that $X$ is on $AY$, and that $E$ is on $BY$-extended. Then $\triangle ACX \sim \triangle ECB$. Let $Z$ be the midpt of $CX$, and $G$ the midpt of $BC$. Then $AZ$ and $EG$ are medians of similar triangles, creating similar sub-triangles: $\triangle ACZ \sim \triangle ECG$. Thus, $A$, $Z$, $F$ are collinear (why?); but $ZF \perp EF$ (why?), so $AF\perp EF$. – Blue Nov 30 '14 at 1:10
• @Blue I'm not sure that X or Y should exist. CE is not tangent to a twin circle unless I'm missing something? – Nick H Nov 30 '14 at 8:02
• I'm referring to the original diagram (not your construction), where tangency to $CE$ is part of what makes a twin circle a twin circle ... isn't it? – Blue Nov 30 '14 at 13:53

## 1 Answer

Restating my comment as an answer with diagrams (and slightly-different labeling) ... Let $X$ be the point of tangency of the twin circle $\bigcirc R$ with $\overline{CE}$, and $Y$ the point of tangency of $\bigcirc R$ with the big semicircle, $\stackrel{\frown}{AB}$.

Introduce $\overline{AY}$ and $\overline{BY}$, which form a right angle at $Y$. Convince yourself that $X$ is on $\overline{AY}$, and that $E$ is on $\overline{BY}$-extended. (This isn't difficult, but it's a bit tedious. See the "footnote" below.) Then $\triangle ACX \sim \triangle ECB$. Now, let $Z$ be the midpoint of $\overline{CX}$, which is in fact the center of $\bigcirc CFX$. (Why?) Then $\overline{AZ}$ and $\overline{EQ}$ are corresponding medians of similar triangles, creating similar sub-triangles: $\triangle ACZ \sim \triangle ECQ$. Consequently, $\angle AXC$ is congruent to $\angle EQC$, which is itself congruent to $\angle FZX$ (why?); this implies that $A$, $Z$, $F$ are collinear. But $Z$ is such that $\overline{ZF}\perp\overline{EF}$ (why?), and we conclude that $\overline{AF}\perp\overline{EF}$.

This is the result from Boas' hint (arrived-at via different means); from here, showing $\overline{AC}\cong\overline{EF}$ is straightforward: $\triangle ACZ$ and $\triangle EFZ$ are similar right triangles with $\overline{CZ} \cong \overline{FZ}$; therefore, the triangles are actually congruent, and $\overline{AC}$ and $\overline{EF}$ are corresponding parts.

Note. For some reason, I'm just not seeing Boas' suggestion to use an inversion to establish $\overline{AF}\perp\overline{EF}$. I don't doubt that his approach is more elegant, especially since the problem is taken from an article celebrating the convenience of inversion. Even without an inversion, I think my argument could use considerable refinement. (It seems like we should be able to take better advantage of $\bigcirc CFX$, which also passes through the intersection of $\bigcirc P$ and $\overline{AY}$.)

Note 2. It's worth mentioning that the radius of the twin circle is given by $\frac{pq}{p+q}$, which is half the harmonic mean of $p$ and $q$, where $p$ and $q$ are the radii of $\bigcirc P$ and $\bigcirc Q$. (Indeed, the construction @Nick used incorporates the "crossed ladders" derivation of the mean.) The fact that this value is symmetric in $p$ and $q$ explains the "twin" part of the name "twin circle": the counterpart of $\bigcirc R$ that is tangent to $\bigcirc P$ has the same size.

Footnote. To show that $X$ is on $\overline{AY}$ and $E$ is on $\overline{BY}$-extended, consider the following, where I've introduced $O$ as the center of the large semicircle: Certainly, $\overline{OY}$ passes through $R$, because $\bigcirc O$ and $\bigcirc R$ are tangent at $Y$. And, since $\overline{RX}\parallel\overline{OA}$, we have that $\triangle OAY$ and $\triangle RXY$ are isosceles triangles with congruent vertex angles; they're similar, so that $\angle OYA\cong\angle OYX$. This (along with the fact that $A$ and $X$ are on the same side of $\overline{OY}$) implies that $A$, $X$, $Y$ are collinear.

Likewise, $B$, $W$, $Y$ are collinear, where $W$ is the diametric opposite of $X$ in $\bigcirc R$. Observe that $E$ lies on a line through $C$ and $X$, and on line through $R$ and $Q$ (not shown above, but shown earlier). $E$ is therefore the center of a dilation between $\bigcirc R$ to $\bigcirc Q$. Necessarily, $E$ is on the line through $W$ and $B$. And, since $\angle QBW \cong \angle RWY$, it follows that $Y$ is collinear with $W$ and $B$ (and $E$).

Of course, going through these details greatly amplifies the imperfection of this approach. I trust that there's a way to streamline my argument to avoid all these collinearity arguments, but I haven't found it.

• Happy to mark this as the answer as it is thoughtful and satisfies much of what I was hoping to find here. Your claim that $\overline{OY}$ passes through R, because $\odot O$ and $\odot R$ are tangent at Y. Is something I would have missed in trying to prove colinearity. – Nick H Nov 30 '14 at 19:59
• It turns out that Cut-the-Knot has a discussion of the twin circles, and a spare! The inversive proof there may relate to the argument hinted-at by Boas for this problem. – Blue Jan 21 '15 at 12:51