The question is as follows:
If a line cuts a triangle into two pieces of equal area, must that line go through the centroid of the triangle? Explain your answer.
I stated yes because we know that the centroid (formed by the intersection of the triangle's medians) create six smaller triangles that will each have $\frac{1}{6}$ of the whole area of the triangle. If you group three of those smaller triangles that have areas of $\frac{1}{6}$ each (which can be seen on either side of a median line), then it will be equal to $\frac{1}{2}$ and the other side will also be $\frac{1}{2}$ of the whole area of the triangle.
Yet I still can't help but second-guess my answer. Is there any other point, other than the centroid, that a triangle can be divided into two polygons of equal area? I have seen posts on StackExchange regarding this problem, but I was not able to understand its complexities for I am just a high-schooler. Any help will be greatly appreciated.
Best Answer
This might not always be true
Consider the following case. Line $EF||BC$ and $AGD$ is perpendicular to $BC$. Assume that area $AEF$ is equal to area $EFCB$.
Let $AE/EB=AF/FC=AG/GD=x:1$. Then, $EF:BC=x:(x+1)$ (why?).
Now, area $AEF$ is half of area $ABC$ (why?) So, $$1/2\cdot AG\cdot EF=1/4\cdot AD \cdot BC\rightarrow 2x^2=(x+1)^2$$ Notice $x\neq 2$
Now, construct a median $AIH$ on $BC$:
If line $EGF$ passes through a centroid, then the centroid must be the point $I$ (why?).
But, since $AI:IH=AG:GD=x\neq2$, hence, line $EGF$ does not pass through a centroid, while still dividing $\triangle ABC$ into two equal areas.
QED.