# $G$ is a point inside triangle $ABC$ such that $[GBC]=[GCA]=[GAB]$, where $[XYZ]$ is the area of $XYZ$. Show that $G$ is the centroid of $ABC$.

Let $$G$$ be a point inside triangle $$ABC$$ such that $$[GBC]=[GCA]=[GAB]$$, where $$[XYZ]$$ is the area of a triagle $$XYZ$$. Show that $$G$$ is the centroid of the triangle $$ABC$$.

My attempt: Since that $$[GBC]=[GCA]=[GAB]$$, so we have $$CG$$, $$AB$$ and $$GB$$, are the $$3$$ medians, so $$G$$ is centroid of $$ABC$$.

I'm not sure about it.

• Perpendicular lengths from other two vertices to median are equal in length (prove it by congruency). Similar for other medians, And hence medians divide triangle into 3 equal areas. All steps are reversible so when Converse is true Aug 18 '20 at 17:23
• ok, I understand, thanks so much! Aug 18 '20 at 17:29

Not really, unless the triangle $$ABC$$ is equilateral.

But this suggests a line of reasoning if you can use affine transformations. We have the following facts:

1. Under an affine transformation, the ratio between two areas is constant.

2. If $$(ABC)$$ and $$(A'B'C')$$ are two non-degenerate triangles, then there exists an affine transformation that maps one onto the other.

Consequently, to solve the problem in general it is sufficient to solve it for an equilateral triangle. And there you have it.

• Nice reasoning and we can use it on the contest. +1 Aug 18 '20 at 17:30

There is an easy proof if you know barycentric coordinates.

Briefly said, barycentric coordinates of a point $$M$$ interior to a triangle $$ABC$$ is the system $$(w_A,w_B,w_C)$$ of $$3$$ numbers (called "weights") to be placed on the vertices $$A,B,C$$ to get a center of mass at $$M$$.

There is an easy way to find these weights (the so-called areal interpretation of barycentric coordinates):

$$w_A=[MBC], \ \ w_B=[AMC], \ \ w_C=[ABM]\tag{1}$$ (https://www.scratchapixel.com/lessons/3d-basic-rendering/ray-tracing-rendering-a-triangle/barycentric-coordinates),

Remark: By their definition, barycentric coordinates are unique, up to a multiplier ; the most usual multiplier is $$1/[ABC]$$ : in this case, we call them normalized barycentric coordinates and their sum is $$1$$.

If all areas $$[GBC]=[GCA]=[GAB]$$ are equal, the normalized barycentric coordinates are $$(1/3,1/3,1/3)$$ : we recognize those of the centroid ; this allows to conclude due to the unicity of barycentric coordinates.

Remark: Barycentric coordinates make sense even when $$M$$ is exterior to triangle $$ABC$$: just consider in (1) that the areas are oriented areas ; for example $$[MBC]$$ is taken as positive if going from $$M$$ to $$B$$, then to $$C$$, one turns with the direct orientation, otherwise $$[MBC]$$ is taken with a negative sign.

Let $$CG\cap AB=\{C_1\}$$, $$BG\cap AC=\{B_1\},$$ $$AG\cap BC=\{A_1\}$$,

$$S_{\Delta AGC}=S_{\Delta AGB}=S_{\Delta CGB}=s$$, $$S_{\Delta GBA_1}=a_2$$ and $$S_{\Delta GCA_1}=a_1.$$

Thus, $$\frac{BA_1}{CA_1}=\frac{a_2}{a_1}=\frac{s+a_2}{s+a_1},$$ which gives $$a_1=a_2$$ and from here $$A_1$$ is a mid-point of $$BC$$.

Can you end it now?

• oh, I understand! Yes, I can to end it. Thanks, I'm learning geometry. Aug 18 '20 at 17:18
• @Ramanujan You are welcome! Aug 18 '20 at 17:20
• I just provided a proof using the very practical barycentric coordinates. Aug 18 '20 at 19:51