I know midline as midbase lol, wlog $BP \ge BM$, as the other case is similar. Using the symedian at the problem we have by angle chase $$\angle DMP-\angle DBC=\angle BDM=\angle CDP \implies (PDM) \; \text{tangent to} \; \Omega$$$$\angle EMP-\angle EBC=\angle BEM=\angle CEP \implies (PEM) \; \text{tangent to} \; \Omega$$So now by Monge d'Alembert on $(PDM),(PEM),\Omega$ we get that the common external tangets and $AP$ are concurrent at $X$. Now consider an inversion with center $X$ that sends $(PDM) \to (PEM)$, clearly $D \to E$ but since $\Omega$ is a circle tangent to both circles + it has 2 fixed points down inversion we have that $\Omega \to \Omega$ and also note that $P \to P$ so this inversion is inversion with center $X$ and radius $P$ so the circle with center $X$ and radius $XP$ is ortogonal to $\Omega$, its clear that $XP=XM$ and now by inversion w.r.t. $\Omega$ we get that $M$ is sent to $A$ but since $M$ lies on a circle ortogonal to $\Omega$ we get that $A$ lies in the circle with center $X$ and radus $XP$ so $XA=XP$ meaning that $X$ is the midpoint of $AP$ so $X$ lies in the A-midbase thus we are done

I checked the Russian wording and corrected the wording from ''median line'' to ''midline''

parmenides51 wrote: I checked the Russian wording and corrected the wording from ''median line'' to ''midline'' The problem is that $AP$ is an arbitrary line so the problem is probably wrong (maybe i did a fakesolve but now checking by geogebra so far everthing i did its correct) @above Oh now i realiced that i missinterpreted midline with median line, sorry, gonna complete my proof Edit: I fully solved the problem now

concurrence point lies

, so also on midline

Note that $DP$ is symmedian in $DBC$ so $\angle PDC = \angle MDB$. Claim $: PMD$ and $PME$ are tangent to $CDB$. Proof $:$ Note that $\angle PMD = \angle MDB + \angle MBD = \angle EDC + \angle CBD = \angle EBD$ with same approach we have $\angle PME = \angle DBE$ Let $O$ be center of $DBC$ and $O_1$ be center of $PMD$ and $O_2$ be center of $PME$. through last claim we have $D,O_1,O$ and $E,O_2,O$ are collinear. Claim $: PO_1OO_2$ is parallelogram. Proof $:$ Note that $\angle O_1OO_2 = \angle DOE = 2\angle DBE = \angle DBE + \angle MDB + \angle MEB = \angle DME = \angle 90 - \angle MDE + \angle 90 - \angle MED = \angle O_1PM + \angle O_2PM = \angle O_1PO_2$ and $PO_1 || EO || O_2O$ Let $O_1O_2$ meet $AP$ at $S$. Note that since $O_1O_2$ is perpendicular bisector of $PM$ then $AS = SP$. Since $O_1O || PO_2$ we have $\frac{SO_1}{SO_2} = \frac{O_1D}{O_2P} = \frac{R_{PDM}}{R_{PEM}}$ so $S$ is where common external tangents of $MDP$ and $MEP$ meet. Since $AS = SP$ so $S$ lies on midline of $ABC$.

Let $S$ be the intersection of the external tangents ;$T,T'$ the tangency points of external tangent with the circles . $S$ is also the ex-similicenter of $(DPM),(PME)$ then $SD.SE=ST.ST'$ more $ST^2.ST'^2=SD.SP.SP.SE$ implies $SD.SE=SP^2$ , let $P'$ the reflection of $P$ in $S$ then $(P,P';D,E)=-1$ but $(P,A;D,E)=-1$ then $A=P'$ which ends the proof .