Wings that cut a chord equally

Draw a chord $PQ$ across a circle and mark its midpoint $M$. Through $M$ draw any two other chords, $AB$ and $CD$, at whatever angles. Their four ends lie on the circle, and joining them crosswise — $A$ to $D$, and $B$ to $C$ — gives two lines that cross the original chord at two points, $X$ and $Y$. The picture looks like a butterfly sitting on the chord: two triangular wings, $AMD$ and $BMC$, meeting at the body $M$.

The theorem is that $MX = MY$. The wings cut the chord at equal distances from its midpoint, whatever the two chords through $M$ are. Nothing in the construction was symmetric. The two chords were drawn at arbitrary angles, the wings are different shapes and different sizes, and one of them can be long and thin while the other is short and fat. The equality is not visible, and it holds every time.

The problem was posed in 1803 by William Wallace in a magazine of mathematical puzzles, and the first known solution, by William George Horner — the Horner of the polynomial method — appeared in 1815. It has since accumulated dozens of proofs: by trigonometry, by coordinates, by projective geometry, by areas, and by inversion, which turns the chords through $M$ into circles and the wings into something else again. It is one of the most proved theorems of elementary geometry, and the reason is that every proof feels like a trick, and nobody is satisfied that the trick is the reason.

This essay gives the proof that uses only what the earlier essays on the inscribed angle built — equal angles on equal arcs, equal products along chords through a point and the law of sines with the circle’s diameter — and then asks what the theorem is really about, which turns out not to be circles at all.

Four hundred butterflies #

Before any proof, it is worth seeing that the claim survives being tested on cases nobody chose.

Each dot is a butterfly with its two chords tilted at random. As the chords tilt, $X$ and $Y$ slide along $PQ$ — towards $M$ when the chords are nearly perpendicular to $PQ$, away from it when one chord lies close to $PQ$ — and the distances range from a small fraction of the chord to most of its half-length. But the dots never leave the diagonal. Whatever $X$ does, $Y$ does the same in mirror image.

That is a striking thing to see, and it should prompt the question a figure cannot answer: why should a construction with no symmetry produce a symmetric result? The circle is symmetric about the perpendicular through $M$, and $PQ$ is perpendicular to that line, but the two chords through $M$ break the symmetry completely. Something in the circle must be restoring it.

What the proof stands on #

The proof needs three facts about circles, each the subject of an earlier essay, and the figure marks the first two.

Equal angles on the same arc. The angle $DAB$, seen from $A$, and the angle $DCB$, seen from $C$, both look at the arc $BD$, so they are equal — each is half the angle that arc subtends at the centre. Likewise the angles $ADC$ and $ABC$ both look at the arc $AC$. In the figure the first pair measures $22.9°$ and the second $37.2°$. So the wing $AMD$ and the wing $CMB$ have the same angles at their outer corners, crosswise: the angle at $A$ equals the angle at $C$, and the angle at $D$ equals the angle at $B$. The two wings are similar triangles.

Equal products along chords through a point. The power of the point $M$ says that any chord through $M$ is cut into two pieces whose lengths multiply to the same number: $MA \cdot MB = MC \cdot MD = MP \cdot MQ$. Since $M$ is the midpoint of $PQ$, that number is $MP^2$, a quarter of the square of the chord’s length. In the figure it is $0.9375$.

The law of sines. In any triangle, each side divided by the sine of the opposite angle gives the same number. Applied to the four small triangles $AMX$, $DMX$, $BMY$ and $CMY$ — the triangles each wing makes with the chord — it turns the distances $MX$ and $MY$ into ratios of sines and lengths.

The combination takes four applications of the law of sines and one of the power of a point, and it is worth doing once in full, because the shape of the calculation is the shape of the theorem. Write $x = MX$, $y = MY$ and $m = MP = MQ$, with $X$ on the side of $Q$ and $Y$ on the side of $P$.

In the triangle $MXA$ the law of sines gives $x / AX = \sin\angle XAM / \sin\angle AMX$, and in the triangle $MXD$ it gives $x / XD = \sin\angle XDM / \sin\angle DMX$. Multiplying the two,

$$\frac{x^2}{AX \cdot XD} = \frac{\sin\angle XAM \cdot \sin\angle XDM}{\sin\angle AMX \cdot \sin\angle DMX}.$$

The same two steps in the triangles $MYB$ and $MYC$ give the same kind of formula for $y^2 / (BY \cdot YC)$, with the angles at $B$, $C$ and at $M$ on the other side. Now the circle does its work twice over. The angles at the circle are equal in pairs by the inscribed-angle theorem: the angle at $A$ in the first wing stands on the arc $BD$, as does the angle at $C$ in the second, and the angle at $D$ stands on the arc $AC$, as does the angle at $B$. The angles at $M$ are equal in pairs because they are vertically opposite: the angle between $MA$ and the chord towards $Q$ is the angle between $MB$ and the chord towards $P$, and similarly for $MD$ and $MC$. So the two right-hand sides are identical, and

$$\frac{x^2}{AX \cdot XD} = \frac{y^2}{BY \cdot YC}.$$

The last step is the power of a point, applied not at $M$ but at $X$ and at $Y$. The chord $AD$ and the chord $PQ$ both pass through $X$, so $AX \cdot XD = PX \cdot XQ = (m + x)(m - x) = m^2 - x^2$; likewise $BY \cdot YC = m^2 - y^2$. Substituting,

$$x^2 (m^2 - y^2) = y^2 (m^2 - x^2), \qquad\text{so}\qquad x^2 m^2 = y^2 m^2,$$

and $x = y$. The terms $x^2 y^2$ cancel from both sides, which is the whole trick, and the cancellation happens because $PX \cdot XQ$ and $PY \cdot YQ$ are both written using the same half-length $m$ — that is, because $M$ is the midpoint. That last step is a hint about what happens when it is not.

Off the midpoint: a law of reciprocals #

Move $M$ along the chord, away from the middle, and repeat the construction. Now the two distances are not equal, and the figure shows how unequal they are.

The open circles, the difference $MX - MY$, scatter: once $M$ is off the middle, how unequal the wings’ cuts are depends on how the chords through $M$ are tilted. The dots, the difference of the reciprocals, do not scatter at all. For every pair of chords,

$$\frac{1}{MX} - \frac{1}{MY} = \frac{1}{MQ} - \frac{1}{MP},$$

where $X$ is the crossing on $Q$’s side of $M$ and $Y$ on $P$’s. The right-hand side depends only on where $M$ sits on the chord, not on the chords through it. This generalisation was published by Leonard Candy in 1896, and it is what the butterfly theorem was all along: a statement that a certain combination of reciprocal distances is fixed by the chord and the point. At the midpoint, where $MP = MQ$, the right-hand side is nought, the reciprocals are equal, and so are the distances.

Seen this way, the symmetry of the butterfly is not a symmetry of the construction. It is the special value of a law that holds everywhere along the chord, at the one place where the law’s constant happens to vanish. That is a common shape for a surprising theorem — an equality that is really an identity with a term that drops out — and it explains why every direct proof feels like a trick: the proofs work hard to establish an equality whose natural form is the reciprocal law.

Candy’s law as a map of the chord #

There is a way to read the reciprocal law that prepares for the projective explanation. Fix the chord $PQ$ and the point $M$, and think of the construction as a machine: choose a point $X$ on the chord, find a pair of chords through $M$ whose wing $AD$ passes through $X$, and read off where the other wing crosses — that is $Y$. The law says the answer does not depend on which pair of chords was used, so the machine is a well-defined map from points $X$ to points $Y$ of the line.

Measured from $M$ with signs, positive towards $Q$, the law reads $1/x + 1/y = c$, where $c$ is the constant $1/MQ - 1/MP$. A map of that form — reciprocal, shift, reciprocal — is a Möbius transformation of the line, and because the equation is symmetric in $x$ and $y$ it has a special property: applied twice, it is the identity. That is no accident of algebra; the roles of $X$ and $Y$ in the construction can be exchanged by swapping the names of the chords. A map that undoes itself is an involution. It swaps $P$ with $Q$, since a wing through $P$ is one whose end is $P$ itself, and the other wing then ends at $Q$. And it fixes $M$, in the limit where both wings close up on the body.

When $M$ is the midpoint, $c = 0$ and the involution is $y = -x$: reflection in $M$. That is the butterfly theorem stated as a property of a map rather than of two lengths, and in that form it is easy to see why it cannot depend on the circle’s size, on the chord’s position, or on how the chords through $M$ are tilted. An involution of a line is determined by two of its swapped pairs; once it swaps $P$ with $Q$ and fixes their midpoint, it is the reflection, and every other pair it swaps is a mirror pair. The question of why the wings are symmetric has become the question of why the construction defines an involution at all — and that is a question about how lines meet a curve of degree two.

The surprising thing: the circle is not needed #

The proof above used the circle three times, through inscribed angles, the power of a point and the law of sines. All three are facts about circles. And yet the butterfly theorem is true on every conic.

On an ellipse the equal-angle argument fails: inscribed angles on an ellipse are not equal, and the power of a point is not constant along its chords in the same way. But the butterfly survives. On a parabola, which is not even closed, the wings still cut the chord equally. The theorem holds on a hyperbola too, and on a pair of lines, which is a degenerate conic.

The reason is that the butterfly theorem is a statement of projective geometry: geometry that keeps only straight lines and their meetings, and forgets lengths and angles. A projection — the shadow cast from a point onto a plane — sends the circle to any conic, sends straight lines to straight lines, and sends the four points $P$, $Q$, $M$ and the point at infinity along the chord to four points with the same cross-ratio. “$M$ is the midpoint of $PQ$” is the statement that $M$ and the point at infinity divide $P$ and $Q$ harmonically, and that statement survives projection. What the theorem says about $X$ and $Y$ — that they are mirror images through $M$ — is likewise a harmonic statement, and so it survives too.

The deepest form of it is Girard Desargues’s involution theorem of 1639: the conics through four fixed points cut any line in pairs of points that are swapped by one and the same involution of the line, a map that undoes itself. The circle, the two chords through $M$ taken as a degenerate conic, and the pair of wings taken as another, all pass through the four points $A$, $B$, $C$, $D$; so the pairs $\{P, Q\}$, $\{M, M\}$ and $\{X, Y\}$ are swapped by a single involution of the line $PQ$. An involution that swaps $P$ with $Q$ and fixes $M$, their midpoint, is the reflection in $M$ — and so it swaps $X$ with $Y$ symmetrically. The circle’s roundness never enters; Pascal’s hexagon theorem lives in the same world for the same reason.

So the three circle facts used in the elementary proof were scaffolding. They are true, and they prove the theorem for circles, but the theorem’s real support is the projective structure that the circle shares with every conic. That is the surprising connection: a theorem about equal lengths, proved with angles and products, turns out to need neither lengths nor angles, and is a fact about how lines meet curves of degree two.

Why the wings must cross the chord at all #

A small detail is worth settling, because the figures depend on it. The two wings, $AD$ and $BC$, join points on opposite sides of $PQ$ — $A$ above and $D$ below, or the other way — and so each wing crosses the line $PQ$ somewhere. Whether it crosses inside the chord, between $P$ and $Q$, depends on the angles; for chords through $M$ that are not too close to $PQ$ itself, both crossings are inside. The four hundred random butterflies were drawn with their chords kept away from $PQ$ for that reason. When a chord through $M$ approaches $PQ$, one wing’s crossing runs out towards $P$ or $Q$ and, in the limit, beyond — and the theorem continues to hold for the line $PQ$ extended, with the signed distances behaving exactly as Candy’s law requires.

The proof’s ingredients drawn, its algebra not #

The proof is described, not drawn. The angle figure marks the equal angles and states the equal products, and those are the ingredients; the algebra that combines them through the law of sines in four triangles is stated in one line. No figure here shows the cancellation happening, and a reader who wants the proof must do the four applications of the law of sines on paper.

The projective argument needs the point at infinity. The cross-ratio and harmonic division that explain why the theorem holds on every conic involve the point where the chord meets the line at infinity, which no picture contains. The conic figure shows that the theorem holds on an ellipse and a parabola by measuring; that it must hold, by projection, is the argument in the text.

Every butterfly drawn has its wings crossing inside the chord. The configurations where a wing crosses $PQ$ outside the circle, or where $M$ is outside the chord, are covered by the signed version of the theorem and are not drawn.

Still open: how many proofs there are #

The butterfly theorem is settled, and the open questions it raises are of a different kind. It has been proved by synthetic geometry, by trigonometry, by coordinates, by projective involutions, by complex numbers, by areas and by inversion, and collections of its proofs run to dozens. Each proof organises the same facts differently, and there is no agreed account of which proof is the explanation — of whether the symmetric statement or Candy’s reciprocal law is the more fundamental, or whether the projective proof, which explains the most, is too far from the picture to count as seeing why. Questions of that kind are not mathematical questions in the narrow sense, and they are how a theorem with dozens of proofs keeps generating new ones.

There are also extensions with open corners. The butterfly has analogues for quadrilaterals inscribed in conics, for chords of quadrics in three dimensions, and for curves of higher degree; for cubic curves the analogous statements involve the group law on the curve, and which of the circle’s theorems lift to higher degree, and in what form, is understood case by case rather than all at once. The next statement in this subject is of a different kind altogether: a constant total hidden in any cutting of a polygon inscribed in a circle into triangles, which was hung on a temple wall in Japan.