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I have attempted to solve this task: Sketch

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.

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Consider the following figure enter image description here

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.

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  • $\begingroup$ Thank you very much, you solved my problem :) I have one question to your answer: Is it metodically correct to assume that a median divides the triangle into two triangles with the same surface area, or it should be proved? $\endgroup$ – ILoveChess Feb 6 '17 at 22:01
  • $\begingroup$ Same base (since the median goes to the mid point) and same height. It´s already proved. $\endgroup$ – Grouper Nov 19 '17 at 22:13
  • $\begingroup$ Beautiful answer! $\endgroup$ – DatBoi Dec 21 '20 at 16:23
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Apply an affine transformation to produce an equilateral triangle. Such a transformation preserves relative areas, and all the little triangles are now the same.

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  • $\begingroup$ Sorry, I don't know anything about affine transformations, it's a high school level problem. $\endgroup$ – ILoveChess Feb 6 '17 at 21:24
  • $\begingroup$ It's just a linear stretch. Worth learning about. $\endgroup$ – Bill Kleinhans Feb 8 '17 at 4:10
  • $\begingroup$ love this strategy! $\endgroup$ – Anirudh Jan 16 '18 at 3:38
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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.

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  • $\begingroup$ The median is not the height of the triangle, why then did decreasing its length three times result in the whole area being reduced three times? $\endgroup$ – ILoveChess Feb 6 '17 at 21:15
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    $\begingroup$ Draw the altitude of the triangle from C to some point T on the base AB. Draw a line through the centroid G parallel to AB. By similar triangles you see height of triangle GAB = 1/3 height of triangle CAB = GD.CD. $\endgroup$ – victoria Feb 6 '17 at 21:25
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    $\begingroup$ @ILoveChess: It is methodically correct not prove the area halving property. However, you can find the very intuitive proof here :en.wikipedia.org/wiki/Median_(geometry) $\endgroup$ – zoli Feb 6 '17 at 22:06

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