Bernhard Elsner, alias MathOMan, posted this exercise in plane Geometry, Theorem about a circle, three chords and a midpoint on January $29$th, $2010.$

"Let $\mathcal{C}$ be a circle, $A,B$ two distinct points on $\mathcal{C}$ and $M$ be the midpoint of the chord $[AB]$. Take two other chords,$[PQ]$ and $[SR]$, that pass through $M$. Let $C$ (resp. $D$) be the intersection of $[AB]$ with $[PS]$ (resp. $[RQ]$). Prove that $M$ is the midpoint of the chord $[CD]$."

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To prove it I've written the following (failed) argument, in the German version of this post (translation of mine):

alt text

The figure is symmetric with respect to $M$: $\overline{AM}=\overline{MB}$, $\overline{PM}=\overline{MU}$, $\overline{RM}=\overline{MW}$, $\overline{QM}=\overline{MV}$, $\overline{QR}=\overline{VW}$, $\overline{SM}=\overline{MT}$. From $\dfrac{\overline{SC}}{\overline{DT}}=\dfrac{\overline{CM}}{\overline{MD}}=1$ follows that $\overline{CM}=\overline{MD}$.

Here is an extract of the author's reply (translation of mine):

"It is not clear that $\overline{QM}=\overline{MV}$. Is the point $V$ on the line $(QM)$ defined by this equality or is $V$ defined as the intersection point of the lines $(QM)$ and $(WC)$? Why are both definitions to give the same point?

Let $C^{\prime }$ be the intersection of $(SP)$ and $(VW)$, and $D^{\prime }$ the intersection of $(TU)$ and $(QR)$. (...)

One still has to show that $C^{\prime }=C$ and $D^{\prime }=D$."

I have agreed with these objections.

Until now no proof has been posted. The author considers that the "proof is not quite simple".

Q. What is the theorem this exercise refers to? Or how does one prove it?

  • 7
    $\begingroup$ The theorem is know as the "Butterfly Theorem", and you can find a few proofs here cut-the-knot.org/pythagoras/Butterfly.shtml $\endgroup$ Aug 12, 2010 at 14:54
  • $\begingroup$ @Jorge Miranda +1, Your comment is also an answer. Thanks! $\endgroup$ Aug 12, 2010 at 16:29
  • $\begingroup$ Should this question be closed? I am not sure, but I will understand if someone who has the power to do it, decides so. I did NOT know this theorem was stated and proved in the Cut The Knot site. $\endgroup$ Aug 12, 2010 at 16:44
  • 1
    $\begingroup$ I posted this question on the meta.math "A question asked in a blog post that I was not aware that it was a theorem stated and proved in another site" meta.math.stackexchange.com/questions/604/… $\endgroup$ Aug 12, 2010 at 17:37
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    $\begingroup$ Instead of just posting a link which may one day die, I'd rather see someone present a summary of the answer here, with proper attribution. I would consider that an ideal response. Of course, other members here may come up with proofs not listed on that site, and those would be great to have here too! @Jorge: In this case, since part of his question is "what is the name of this theorem," your comment is a reasonable answer. If he had only been interested in the proof, maybe not. $\endgroup$
    – Larry Wang
    Aug 12, 2010 at 17:50

1 Answer 1


The theorem is known as the "Butterfly Theorem".

One fairly simple proof is the following (from Selected Problems and Theorems of Elementary Mathematics by D. O. Shklyarsky, N. N. Chentsov and I. M. Yaglom).

Let O be the center of the given circle. Since OM ⊥ CD, in order to show that CM = MD, we have to prove that ∠COM = ∠DOM.

Drop perpendiculars OK and ON from O onto PS and QR, respectively. Obviously, K is the midpoint of PS and N is the midpoint of QR. Further, ∠PSR = ∠PQR and ∠QPS = ∠QRS,

as angles subtending equal arcs. Triangles SPM and QRM are therefore similar, and SP/SM = QR/QM, or SK/SM = QN/QM. In other words, in triangles SKM and QNM two pairs of sides are proportional. Also, the angles between the corresponding sides are equal. We infer that the triangles SKM and QNM are similar. Hence, ∠SKM = ∠QNM.

Now, have a look at the quadrilaterals OKCM and ONDM. Both have a pair of opposite straight angles, which implies that both are inscribable in a circle. In OKCM, ∠SKM = ∠COM. In ONDM, ∠QNM = ∠DOM. From which we get what we've been looking for: ∠COM = ∠DOM.

  • $\begingroup$ I checked for typos and found none. I also set this answer as accepted. Above I've written "Inspired by Kaestur's comment, let's now see if someone writes a proof from another source or establishes a new one." $\endgroup$ Aug 14, 2010 at 12:29
  • $\begingroup$ I inform you that there is a ongoing discussion concerning verbatim quotation, copy, etc. Please take into consideration the outcome of it. meta.math.stackexchange.com/questions/656/… To me is quite enough to have the theorem's name and a link to the source, as you did in your 1st comment and in the 1st paragraph of your answer. $\endgroup$ Aug 15, 2010 at 0:15
  • $\begingroup$ Update: concerning this the first comment in the meta thread states: "There is nothing wrong at all with copying a small proof like that [this] with attribution" and considers this a non-issue. I will not comment further on it, unless the outcome is very different. Apologies if I am bothering you. $\endgroup$ Aug 15, 2010 at 10:48

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