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?