parmenides51 wrote: A line passing through the center $M$ of the equilateral triangle $ABC$ intersects sides $BC$ and $CA$, respectively, in points $D$ and $E$. Circles circumscribed around triangle $AEM$ and $BDM$ intersects, besides point $M$, also at point $P$. Prove that the center of the circle is circumscribed around the triangle $DEP$ lies on the line of the segment $AB$. I guess the question should be: A line passing through the center $M$ of the equilateral triangle $ABC$ intersects sides $BC$ and $CA$, respectively, in points $D$ and $E$. Circumcircles of triangle $AEM$ and $BDM$ intersects, besides point $M$, also at point $P$. Prove that the center of circumcircle of triangle $DEP$ lies on the perpendicular bisector of the segment $AB$.

yes, thanks for the correction (it said axis of a segment and it meant perpendicular bisector, not a line as I misunderstood)

Let $O$ be the circumcenter of $\triangle DEP$. Note that it suffices to prove that $C-M-O$ are collinear, so we are going to prove that. With the cyclic quadrilaterals given we get $\angle MAE=\angle MPE=30^{\circ}$ and $\angle MBD=\angle MPD=30^{\circ}$, then $\angle DPE= \angle MPD+\angle MPE=60^{\circ}$, and since $O$ is circumcenter of $\triangle DEP$, $\angle DOE=120^{\circ}$, $\implies DOEC$ cyclic. Now since $OD=OE$ we get that $OC$ is lies in the bisector of $\angle DCE$, and clearly $M$ does as well, and we are done.

Attachments:

After noticing $P$ belongs to the circumcircle of $ABC$, invert through $P$ and it will be easy to prove that $M'$, $O'$, $P$ and $C'$ is an isosceles trapezoid symmetric with respect to $D'E'$where $O$ is the center of $ABC$. Since, it is cyclic then $M$, $O$ and $C$ are collinear as desired.