This is what I found several years ago when I was in middle school:
Suppose we have a circle on a plane and arbitrarily choose four different points on the circle, say $P,A,B,C$. Then draw three circles with center $A,B,C$ and radius $|PA|, |PB|, |PC|$, respectively. The three new circles meet at three crossover points besides $P$, say $D, E, F$.
I was amazed that $D, E, F$ are on the same line. Could someone prove this?
On
The points $D$, $E$, $F$ are reflections of $P$ in the side-lines of $\triangle ABC$, so the midpoints of segments $PD$, $PE$, $PF$ are the feet of the perpendiculars from $P$ to those side-lines. Those feet are "known" to be collinear, determining the Simson line.
Consequently, the original line is the dilation of the Simson line in $P$ with scale factor $2$.
Invert around point P. Let $X'$ denote the image of the point $X$ under this inversion. We find that $(A'B'C')$ is a straight line, and the circles with centers $A,B,C$ and passing through $P$ are mapped to perpendicular bisectors of the segments $PA',PB',PC'$ respectively, call the perpendicular bisectors $p_a,p_b,p_c$ respectively. Now let the midpoint of the segments $PA',PB',PC'$ be $R,S,T$ respectively. obviously $R,S,T$ are collinear and $R-S-T$ is parallel to $A'-B'-C'$. Let $p_a\cap p_b=D',p_b\cap p_c=E', p_a\cap p_c=F' $. Then we have, in the geometry of the attached figure, that $\angle F'PR= \angle F'TR=\angle E'TS=\angle E'PS\implies \angle E'PF'=\angle SPR +\angle E'PS-\angle F'PR=\angle SPR$. also note that $\angle E'D'F'=\angle SD'R=180^{\circ}-\angle SPR$ because $D'SPR$ is cyclic. Hence $\angle E'PF'+\angle E'D'F'=\angle SPR+ 180^{\circ}-\angle SPR=180^{\circ}$ Hence $PE'D'F'$ is cyclic. Inverting back, we see that $D-E-F$ is a straight line as desired. Other configurations are handled similarly.