Proving a point inside a triangle is the centroid using ratios. Let P be a point inside triangle ABC. Let the ray AP intersect with BC at A’, BP intersect with AC at B’, and CP intersect with AB at C’.
Prove that if AP/PA’=BP/PB’=CP/PC’ then P is the centroid of ABC.

 A: We use barycentric coordinates. Let $P=(x,y,z)$. Then
$$ \frac{AP}{PA'}=\frac{y+z}{x}$$
and cyclic permutations so
$$\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}.$$
Multiplying through by $xyz$ gives
$$y^2z+yz^2=xz^2+x^2z=x^2y+xy^2$$
The first equation is equivalent to
$$z(y-x)(x+y+z)=0$$
and since $P$ is inside the triangle we have $x,y,z>0$ so, necessarily, $x=y$. Similarly, we can get $y=z$ from another pair of equations, hence $x=y=z$ and $P$ is the centroid.
A: Another solution using areas: The condition is equivalent to
$$\frac{AA'}{PA'}=\frac{BB'}{PB'}=\frac{CC'}{PC'}$$
And hence this gives the area ratios
$$\frac{[BCP]}{[BCA]}=\frac{[CAP]}{[CAB]}=\frac{[ABP]}{[ABC]}.$$
On the other hand, however, $[BCP]+[CAP]+[ABP]=[ABC]$ and so each of these ratios must be $1/3$. In turn, each of the ratios of lengths above equals $3$ and hence $P$ divides each of $AA', BB'$ and $CC'$ in the ratio $2:1$. Thus, $P$ is the centroid.
[To see why $P$ must now be the centroid, you can again use areas. Since $AP/A'P=2$, we have $[BAP]/[A'BP]=2$. On the other hand, we already established $[BAP]/[ABC]=1/3$ and so $[A'BP]=1/6[ABC]$. Similarly, $[CA'P]=1/6[ABC]$ and thus $[A'BP]=[CA'P]$, which means $CA'=A'B$ and $A'$ is the midpoint of $BC$. Similarly, $B'$ and $C'$ are the midpoints of the other two sides.]
