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!

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.)
$$\underbrace{\begin{array}{ccc}
\color{red}{\alpha :=\dfrac12(1+\lambda)} &\quad\stackrel{\displaystyle\star}{\to}\quad& \alpha' =\dfrac12(1-\lambda) \\[8pt]
\downarrow{\displaystyle\star\star} && \downarrow{\displaystyle\star\star}  \\[8pt]
\beta'=\dfrac{1}{2(1+\lambda)} && \gamma=\dfrac{1}{2(1-\lambda)} \\[8pt]
\downarrow{\displaystyle\star} && \downarrow{\displaystyle\star}  \\[8pt]
\beta=\dfrac{1+2\lambda}{2(1+\lambda)} && \gamma'=\dfrac{1-2\lambda}{2(1-\lambda)}
\end{array}}_{\displaystyle\phantom{\star\star\,}\downarrow\;\star\star} \\
\phantom{\quad\to\quad \lambda=0}\frac{1-4\lambda^2}{1-\lambda^2}=1 \quad\to\quad \color{red}{\lambda=0}$$
