We know that connecting the centroid of a triangle to the vertices gives us three smaller triangles of equal area. That is, if is the centroid of , then the triangles , and are equal in area.
In this article, we are interested in the converse:
If is a point inside such that , and have equal area, then prove that must be the centroid.
Proof (I)
Extend to meet in , as shown in the figure below.
and have a common vertex and their bases and lie on the same line. Therefore, they have the same height (i.e. distance of from ) and their areas are proportional to their bases:
and have a common vertex and their bases and lie on the same line. Therefore, they have the same height (i.e. distance of from ) and their areas are proportional to their bases:
But it is given that . Therefore, . This means is the median.
Similarly, and when extended bisect and respectively. This proves that is the centroid of .
Proof (II)
Extend to meet in and to meet in , as shown in the figure below.
It is given that .
and have a common vertex and their bases and lie on the same line. Therefore, they have the same height and their areas are proportional to their bases:
Similarly,
Similarly,
Consider and :
In , is parallel to base . . This means that is the midpoint of and is a median.
Similarly, is also a median. Therefore, is the centroid of .
Proof (III)
(Proof by contradiction)
Let
be the centroid of
.
.
Assume that is a point distinct from the centroid .
It is given that .
and have the same area (equal to ). As they have the same base and their respective vertices and lie on the same side of line , they must have the same height. This means, that points and are at the same distance from line . Therefore, is parallel to .
Similarly, we can show that is also parallel to .
intersects at . As is parallel to , must intersect .
Now, we have one line which is both parallel to and intersects . This is a contradiction. As our assumption leads to a contradiction, our assumption must be incorrect. Therefore, must be the centroid of .