How to prove the midpoint of a chord is also the midpoint of the line segment defined by the points of intersection of other two chords with it? 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 $[CD]$."

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

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?
 A: 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.
