Checking angles, we see $\triangle EFP$ is equilateral. The 30-60-90 triangle ratio gives $\frac{PE}{EB}=\frac{EF}{EB}=\frac{\sqrt3}2$, so $\angle EBG$ is constant because its tangent is constant. Set D = C for easy check that this passes in midpoint of $AM$

FalloutBoy wrote: The 30-60-90 triangle ratio gives $\frac{PE}{EB}=\frac{EF}{EB}=\frac{\sqrt3}2$ Maybe I am being dumb, but how is $\frac{PE}{EB}=\frac{\sqrt{3}}{2}$? Based on the diagram that I have, $\triangle{PEB}$ cannot be 30-60-90 as both $\angle{PBE}$ and $\angle{BPE}$ are less than $60^{\circ}$. EDIT: Ah, okay that makes sense. Thanks @below

No, $PE=EF$ and $EFB$ is the 30-60-90, apologies

FalloutBoy wrote: Checking angles, we see $\triangle EFP$ is equilateral. The 30-60-90 triangle ratio gives $\frac{PE}{EB}=\frac{EF}{EB}=\frac{\sqrt3}2$, so $\color{red}\angle EBP$ is constant because its tangent is constant. Set D = C for easy check that this passes in midpoint of $AM$ small typo, I've fixed it.

Thank you! I drew a diagram and labeled that point G sorry

Let $\{O\}=AM \cap BP$ Using Thales: $\dfrac{OM}{PE}=\dfrac{BM}{BE}$ but $BE=FE \cdot \dfrac{2}{\sqrt{3}}; FE=PE$, so $\dfrac{OM}{BM}=\dfrac{\sqrt{3}}{2} \Leftrightarrow AM=2\cdot OM$

As already pointed out, $\triangle FEP$ is equilateral; since $\angle FEP=\angle EFP=\angle EBF$, we infer $BP$ is symmedian of $\triangle BEF$. As $EF$ and $AM$ are antiparallel w.r.t. $AB, BC$, clearly it is median of $AM$. Remark: it was enough $AC=BC$ and $AM$ altitude. Best regards, sunken rock

Let's move the point $D$ on the $BC$ at a constant speed. Then the point $E$ moves on the $AC$ at a constant speed as a projection of point $D$ onto $BC$. Analogously we can get that points $F$ and $G$ move on the $AB$ and $AC$ at a constant speed. And now we want to say that the point $P$ moving in a straight line. This point is determined by the perpendiculars recovered from the points $E$ and $G$ from $BC$ and $AC$. And since points $E$ and $G$ move on the $BC$ and $AC$ at a constant speed, then the point $P$ moves on some line. So, if we prove the problem for two distinct points $D$ ($D_1$ and $D_2$), then we are done since the locus of $P$ it is line passing through the middle of $AM$ and therefore it is line passing through point $B$, so $BP$ bisects $AM$ for any point $D$ on the $AC$. Proof: 1) $D$=$A$. It is trivial. 2) $D$ on the segment $AC$ s.t. $3AD=AC$. Then $3CE=BC$, $3BF=AB$, $3AG=AC$. Then $D=G$. Then $P=D=G$. And the rest is trivial by Menelaus Theorem in triangle $AMC$ and cutting line $BD$.

Dexenberg wrote: Let $\{O\}=AM \cap BP$ Using Thales: $\dfrac{OM}{PE}=\dfrac{BM}{BE}$ but $BE=FE \cdot \dfrac{2}{\sqrt{3}}; FE=PE$, so $\dfrac{OM}{BM}=\dfrac{\sqrt{3}}{2} \Leftrightarrow AM=2\cdot OM$ How do you know that $\triangle BOM \sim \triangle BPE$?