Prove that three medians divide a triangle into 6 triangles with equal surface area.

Question

I have attempted to solve this task:

I have drawn the medians and described the lengths of the segments. I also noticed that triangles ADC and BCD share the same height, thus

$S_{total}= xh$
Whereas the surface of triangles ACD and BCD
$S= xh/2$
And so I have already made it so far as to get the halves. But I haven't got a faintest idea how to get to the sixths.
I would be most grateful if you gave your answer in simple terms. Geometry is my Achilles's heel.

Answer 1

Consider the following figure

The letters in the little triangles stand for the surface areas of the respective triangles. Identical letters denote identical areas of triangles having the same base lengths and the same heights.

We are talking about medians, so we have further equations:

$$2a+b=2c+b.$$

From here, it follows that $c=a$.

Then, we have

$$2a+c=2b+c.$$

As a result $b=a$.

So, $$b=a=c.$$

And this is what we were to prove.

Answer 2

Let G be the centroid .Let $ \Delta(ABC) = a$. Now $\Delta(ACD) = a/2$. Also G divides the median in 2:1 ratio. So $ \Delta(AGD) = a/2.3 = a/6$. Thus each of the triangles have area $a/6$. Hence proved.

Answer 3

Apply an affine transformation to produce an equilateral triangle. Such a transformation preserves relative areas, and all the little triangles are now the same.