Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversely proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever).
For example consider a triangle $ABC$ with its three medians drawn, with the intersection points being $D, E, F,$ corresponding to $AB, BC,$ and $AC$ respectively. Thus, if we label point $A$ with a weight of $1$, $B$ must also have a weight of $1$ since $A$ and $B$ are equidistant from $D$. By the same process, we find $C$ must also have a weight of 1. Now, since $A$ and $B$ both have a weight of $1$, $D$ must have a weight of $2$ (as is true for $E$ and $F$). Thus, if we label the centroid $P$, we can deduce that $DP:PC$ is $1:2$ - the inverse ratio of their weights.
Note that using this method, the location of centroid (ie $P$ being on $CD$ dividing it in the ratio of $1:2$) is not dependent on how we calculate it. Instead of finding $D$ on $AB$, if we found the midpoint $E$ on $BC$, and then located $P$, it would still end up being the same.
This nicely matches the facts about centroid being the point at which the three medians concur. However, why does placing of weights work? It is strange that at any one time, we look at points which are on one line. But in the end, it results in true facts about points in two dimensions.
So why does it work? Is there a way to rigorously prove that it works? Please help me with this question.
In Physics, the center of mass of a set $m_i$ of $n$ point masses situated at positions $\vec{x}_i$, is located by definition at $$ \vec{x}_G={\sum_{i=1}^nm_i\vec{x}_i\over\sum_{i=1}^nm_i}. $$ That is, the position of the center of mass is the weighted average of the positions of the single points, the weight being the mass of each point.
From this definition it follows, for instance, that the center of mass $Q$ of two point masses $m_1$ and $m_2$, located at points $P_1$ and $P_2$, lies on segment $P_1P_2$ and: $m_1 P_1Q=m_2 P_2Q$. If $m_1=m_2$, in particular, $Q$ is the midpoint of $P_1P_2$
Suppose now you divide the set of points into two (or more) subsets and label the points so that the first subset corresponds to indices $1\dots m$ and the second to indices $m+1\dots n$. The centers of mass of the two subsets are located at $$ \vec{x}_A={\sum_{i=1}^m m_i\vec{x}_i\over\sum_{i=1}^m m_i} \quad\text{and}\quad \vec{x}_B={\sum_{i=m+1}^n m_i\vec{x}_i\over\sum_{i=m+1}^n m_i}. $$ If we assign to those subsets a mass equal to the sum of their constituent masses: $m_A=\sum_{i=1}^m m_i$ and $m_B=\sum_{i=m+1}^n m_i$, then you can easily check that the following equality holds: $$ \vec{x}_G={m_A\vec{x}_A+m_B\vec{x}_B\over m_A+m_B}. $$ In other words: you get the same center of mass if you consider each subset as if it were a single mass point, located at the center of mass of the subset and with a mass equal to the sum of the masses in the subset.
Suppose now you assign the same (unit) mass to the three vertices of a triangle $ABC$. By the above result, we can consider at first the center of mass $M$ of side $AB$ (which is the midpoint of $AB$): the center of mass $G$ of the triangle is then located on mediam $CM$ and $CG=2MG$.
But the same can be repeated by considering the centers of mass (midpoints) of the other two sides: we then obtain that point $G$ is common to the three medians, and it divides each median into two parts, one the double of the other. We thus proved the existence of the centroid of triangle $ABC$.