In a triangle ABC, if certain areas are equal then P is its centroid Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at
$D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$, have equal areas, prove that $P$ is the centroid of
$\triangle ABC$.
I am trying to do with Ceva's theorem: $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$ and also with the result $\frac{AP}{PD}=\frac{AF}{FB}+\frac{AE}{EC}$ but having some difficulties. Please give any hint.


Let us give name $a$ for the area of triangles with common area and $x,y,z$ for the other areas. We have (using the following result: the areas of triangles sharing a same altitude with collinear bases are in the ratio of the lengths of these bases):
$$\begin{cases}\dfrac{EA}{EC}&=&\dfrac{x}{a}&=&\dfrac{a+z}{a+y}\\ \dfrac{DC}{DB}&=&\dfrac{y}{a}&=&\dfrac{a+x}{a+z}\\
\dfrac{FB}{FA}&=&\dfrac{z}{a}&=&\dfrac{a+y}{a+x}\end{cases}\tag{1}$$
By taking new variables
$$X=\dfrac{x}{a}, \ \ Y=\dfrac{y}{a}, \ \ Z=\dfrac{z}{a}$$
(1) becomes :
$$\begin{cases}X&=&\dfrac{1+Z}{1+Y}\\ Y&=&\dfrac{1+X}{1+Z}\\
Z&=&\dfrac{1+Y}{1+X}\end{cases}\tag{2}$$

*

*The product of these 3 equations (2) gives

$$\underbrace{XYZ}_P=1 \tag{3}$$


*Expanding the equations of (2) gives

$$\begin{cases}
1+Z&=&X+XY\\
1+X&=&Y+YZ\\
1+Y&=&Z+ZX
\end{cases} \tag{4}$$
Adding equations (4) gives:
$$\underbrace{XY+YZ+ZX}_R=3 \tag{5}$$
Multiplying equations in (4), by $Z$, $X$ and $Y$ resp., we get:
$$\begin{cases}
Z+Z^2&=&XZ+XYZ\\
X+X^2&=&YX+XYZ\\
Y+Y^2&=&ZY+XYZ
\end{cases} \tag{6}$$
adding them and setting $S:=X+Y+Z$, we get:
$$S+(S^2-2R)=R+P$$
Taking into account (3) and (5), we get:
$$S+S^2-12=0, \tag{7}$$
a quadratic whose unique positive root is $S=3$.
Therefore, due to Vieta's relationships, $X,Y,Z$ are roots of the third degree equation in $T$ :
$$T^3-ST^2+RT-P=0 \ \iff \ T^3-3T^2+3T-1=0 \ \iff \ (T-1)^3=0\tag{8}$$
giving the triple root $X=Y=Z=1$. Otherwise said:
$$x=y=z=a\tag{9}$$
As a consequence, the areas of triangles $MAB, AMC, ABM$ are identical with common value $2a$.
Therefore, the barycentric coordinates of $M$ are $([MAB]/[ABC], [AMC]/[ABC], [ABM]/[ABC])=(1/3,1/3,1/3)$  characterizing the centroid of $ABC$.
Edit: I just discovered this similar question If three cevians are concurrent at a point and form triangles of equal area, the point is the centroid with an interesting answer by @almagest whose reasoning is partly the same as mine, with a final clever use of the Arithmetic/Geometric means inequality.
