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.
 A: 
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.
A: This is not a trivial geometry problem... Let $x=AF/FB, y=BD/DC, z=CE/EA$, then we have $xyz=1$. Furthermore, we can calculate that the area of $AFB$ is $\frac{x}{(1+z+yz)(1+x)}$. Similarly we can calculate two others, so, we have
$$\frac{x}{(1+z+xz)(1+x)}=\frac{y}{(1+x+yx)(1+y)}=\frac{z}{(1+y+zy)(1+z)}$$
Notice that $(1+z+xz)=(xyz+z+xz)=z(xy+1+x)=z(xy+xyz+x)=zx(y+yz+1)$, so we have
$$\frac{x}{(1+x)}=\frac{yz}{(1+y)}=\frac{z^2x}{(1+z)}$$
So we have $x(1+z)=z^2x(1+x)$ and $x(1+y)=yz(1+x)$. Substitute $y=1/xz$, we have $x(1+1/xz)=z/(xz)(1+x)$, or, $x(1+xz)=z(1+x)$. This yields a linear equation of $z$, that is, $z=\frac{x}{1+x-x^2}$. Substitute it in, we have
$$(1+\frac{x}{1+x-x^2})=(\frac{x}{1+x-x^2})^2(1+x)$$
Or,
$$(1+2x-x^2)(1+x-x^2)=x^2(1+x)$$
Expanding, we have
$$x^4-4x^3-x^2+3x+1=0$$
Factorizing, we have
$$(x-1)(x^3-3x^2-4x-1)=0$$
One solution is $x=1$, the other positive solution is located $x>4$, where $z=\frac x{1+x-x^2}<0$ (which means $P$ is outside), contradiction. Therefore, the only valid solution is $x=y=z=1$.
A: I will use barycentric coordinates, an excellent reference is
Barycentric Coordinates for the Impatient, Max Schindler, Evan Chen
and it is by chance compact, and easily accesible/readable. We will use the formula in Theorem 10 in it.
Let $P$ be $P=(x,y,z)$ in barycentric coordinates w.r.t. $\Delta ABC$, $x,y,z >0$, $x+y+z=1$.
We may and do assume that the area of the given triangle is normed to $[ABC]=1$.
Then the area $[AFP]$ of $\Delta AFP$ is computed using
$$
\begin{aligned}
A &= (1,0,0)\ ,\\
P &= (x,y,z)\ , \\
F &= [x:y:0]=\left(\frac x{x+y},\frac y{x+y},0\right)\ ,\text{ so}\\
[AFP]
&=
\begin{vmatrix}
1 & 0 & 0\\
\frac x{x+y} &\frac y{x+y} & 0\\
x & y & z
\end{vmatrix}
=
\frac 1{x+y}
\begin{vmatrix}
1 & 0 & 0\\
x & y & 0\\
x & y & z
\end{vmatrix}
=
\frac {yz}{x+y}
=
\frac {xyz}{x(x+y)}
\ .
\end{aligned}
$$
The other two areas are obtained by cyclic permutation of the letters $x,y,z$ (taken in this order).
Then the given equality of areas becomes:
$$
\frac {xyz}{x(x+y)} =
\frac {xyz}{y(y+z)} =
\frac {xyz}{z(z+x)} \ .
$$
So $x(x+y)=y(y+z)=z(z+x)$. Let $s$ be the common value of this expression, and let us obtain some further relation, here with all details...
$$
\begin{aligned}
s &= x(x+y) = x(1-z) = x - xz\ ,\\
s &= y(y+z) = y(1-x) = y - yx\ ,\\
s &= z(z+x) = z(1-y) = z - zy\ ,\\[2mm]
3s &= 1 - (xy+yz+zx)\ ,\\
 s &= ys + zs + xs = y(x-xz)+z(y-yx) +x(z-zy)\\
   &= (xy+yz+zx) -3xyz\ ,\\[2mm]
s(1-3s) &= s(yz+zx+xy) =\sum yz(x-xz)= 3xyz - xyz = 2xyz\ ,\\
3s(1-3s) &= 6xyz = 2(xy+yz+zx) - 2s = 2(1-3s)-2s\ ,
\end{aligned}
$$
and we obtain an equation of degree two in $s$, which is
$$
0 = (2-3s)(1-3s) - 2s =2-11s + 9s^2=(1-s)(2-9s)\ .
$$
Since $s=x(x+y)<1$, we only accept $s=2/9$. The simple elementary polynomials in $x,y,z$ are
$$
\begin{aligned}
e_1 &=x+y+z=1\ ,\\ 
e_2 &=xy+yz+zx=1-3s=\frac 13\ ,\\
e_3 &=xyz=\frac 12s(1-3s)=\frac 1{27}\ ,
\end{aligned}
$$
so $x,y,z$ are (Vieta) the roots of the polynomial in $T$
$$
T^3 - T^2+\frac 13T -\frac 1{27}=\left(T-\frac 13\right)^3\ ,
$$
so the point $P$ is $P=(x,y,z)=\left(\frac 13,\frac 13,\frac 13\right)=[1:1:1]$,
the centroid.
