The problem of congruent areas in a triangle. A problem was posed in front of me and I couldn't solve it after multiple attempts--
Consider any triangle and 3 concurent cevians are drawn from each of its 3 points . Now the figure formed has 6 sub triangles - if the areas of 3 alternate sub triangles are equal then prove that the point of concurrence is the Centriod.
 
 A: 
Assume that $\frac{BP_A}{P_A C}=\lambda$ and $\frac{CP_B}{P_B A}=\mu$. By Ceva's theorem $\frac{AP_C}{P_CB}=\frac{1}{\mu\lambda}$. Moreover:
$$ [BPP_A] = \frac{BP_A}{BC}[BPC]=\frac{BP_A}{BC}\cdot \frac{CP_B}{CA}[ABC] $$
hence
$$ [BPP_A]=\frac{\lambda}{\lambda+1}\cdot \frac{\mu}{\mu+1}[ABC].$$
In a similar way you may compute $[CPP_B]$ and $[APP_C]$ in terms of $\lambda,\mu,[ABC]$ and check that from $[BPP_A]=[CPP_B]=[APP_C]$ it follows that $\mu=\lambda=1$, i.e. $P\equiv G$.
A: Let $P$ be the intersection point inside $\triangle{ABC}$.
Let $D$ be the intersection point of $AP$ with $BC$, and let $E,F$ be the point on $AP$ such that $AE\perp BE, AF\perp CF$ respectively.
Since $\triangle{APB}=\triangle{APC}$, we have $BE=CF$ from which we know that $\triangle{BED}$ and $\triangle{CFD}$ are congruent, and so $BD=CD$. Hence, we know that $AP$ is the median.
Also, from $\triangle{PBD}=\triangle{PCD}$, we have that
$$\triangle{PBC}=2\triangle{PBD}.$$
From $\triangle{PAB}=\triangle{PBC}$, we have that
$$\triangle{PAB}=2\triangle{PBD}$$
from which $AP:PD=2:1$ follows.
It follows from this that $P$ is the centroid.
