Triangle separated in 4 triangles of equal area I have a triangle $ABC$ with all angles less than $90$ degrees. Points $M$, $N$ and $P$ are situated on $BC$, $AC$ and $AB$. The areas of $APN$, $BPM$, $NCM$ and $PMN$ are all equal.
I want to prove that $M$, $N$, $P$ are the midpoints of $BC$, $AC$ and $AB$.
I couldn't get anything out of the area formulas. I don't see any similar triangles either. I think it could be useful to first prove that $BC||PN, AC||PM$ and $AB||MN$, but I don't know how to do that.
Any ideas/hints will be apreciated. Thanks!
 A: Let $\alpha$, $\alpha'$, $\beta$, $\beta'$, $\gamma$, $\gamma'$ be the ratios of the subsegments to the sides of the triangle, so that we have the image shown:

Necessarily, we must have
$$\alpha+\alpha'=\beta+\beta'=\gamma+\gamma'=1\tag{$\star$}$$ $$\alpha\beta'=\beta\gamma'=\gamma\alpha'=\frac14\tag{$\star\star$}$$
(The first relation recognizes that each pair of subsegments comprises a complete side; the second follows from each of the subtriangles $\triangle APN$, $\triangle BNP$, $\triangle CNM$ having one-quarter the area of $\triangle ABC$. (Can you see why?))
We could slog through a solution to the non-linear system to find that every value is $1/2$. Actually, this approach isn't as daunting as it sounds; nevertheless, there's a better way ...
We assign $\alpha := \frac12(1+\lambda)$ for some $\lambda\neq\pm1$, and then argue as below to show that $\lambda=0$, obtaining the result.
(Note: "$\star$" and "$\star\star$" indicate that an implication follows from the corresponding relation above.)
$$\underbrace{\begin{array}{ccc}
\color{red}{\alpha :=\dfrac12(1+\lambda)} &\quad\stackrel{\displaystyle\star}{\to}\quad& \alpha' =\dfrac12(1-\lambda) \\[8pt]
\downarrow{\displaystyle\star\star} && \downarrow{\displaystyle\star\star}  \\[8pt]
\beta'=\dfrac{1}{2(1+\lambda)} && \gamma=\dfrac{1}{2(1-\lambda)} \\[8pt]
\downarrow{\displaystyle\star} && \downarrow{\displaystyle\star}  \\[8pt]
\beta=\dfrac{1+2\lambda}{2(1+\lambda)} && \gamma'=\dfrac{1-2\lambda}{2(1-\lambda)}
\end{array}}_{\displaystyle\phantom{\star\star\,}\downarrow\;\star\star} \\
\phantom{\quad\to\quad \lambda=0}\frac{1-4\lambda^2}{1-\lambda^2}=1 \quad\to\quad \color{red}{\lambda=0}$$
A: Let after an affine transformation ( https://en.wikipedia.org/wiki/Affine_transformation )
$\Delta ABC$ goes to equilateral $\Delta A'B'C'$ and points $P,$ $M$ and $N$ go to $P'$, $M'$ and $N'$.
Thus, since $$S_{\Delta A'P'N'}=S_{\Delta B'M'P'},$$ we see that $\Delta A'P'N'\cong\Delta B'M'P'.$
Now, let $A'B'=1$ and $A'P'=x.$
Thus, $B'M'=x$ and we obtain:
$$\frac{1}{2}x(1-x)\sin60^{\circ}=\frac{1}{4}\cdot\frac{\sqrt3}{4},$$ which gives $x=\frac{1}{2},$ $P'$, $M'$ and $N'$ are mid-points of $A'B'$, $B'C'$ and $A'C'$ and from here  $P$, $M$ and $N$ are mid-points of $AB$, $BC$ and $AC$.
Let $A'P'=x$, $B'M'=y$, $C'N'=z$ and $A'B'=1$.
Thus, since $$S_{\Delta A'P'N'}=S_{\Delta B'M'P'}=S_{\Delta C'N'M'},$$ we obtain:
$$x(1-z)=y(1-x)=z(1-y)$$ and from here $$x=y=z.$$
A: Plus these two rather complicated solutions this problem can also have a simple solution:
Draw two lines from A and B parallel with CB and AC respectively. Mark their intersection as D, a parallelogram ADBC will be constructed. Extend MP and NP  to intersect AD and BD at E and F respectively. Triangles PMB and PAE are similar, therefor triangle ADB can be considered as a kind of Affine transform of ABC. Hence $\triangle PMB=\triangle APE$ which results in $\triangle ANP=\triangle APE$. These two triangles have common base AP and two common vertices A and P so their altitudes must be equal. This gives this result that NP must be parallel with AD and BC. Similarly we can show that MP is parallel with AC.This what you were looking for.
