Area of a triangle The following problem in elementary geometry was proposed to me. As a mathematical analyst, I confess that I can't solve it. And I have no idea of what I could do. Here it is: pick a triangle, and draw the three mediana (i.e. the segments that join a vertex with the midpoint of the opposite side). Use the three segments to construct a second triangle, and prove that the area of this triangle is $3/4$ times the area of the original triangle.
Any help is welcome.


If you don't understand vector that well then you can use this proof
Assume that three medians BD,AE and CF are p,q,r respectively and the triangle made by the medians is $\triangle {X}$
Now we draw parallelogram $ GCIA $. Area of $ \triangle {GAC}= \frac {1}{3} \triangle {ABC} $ ( As the medians divide a triangle into six congruent triangles. And there is two of them in $ \triangle {GAC} $ ).
Now $\triangle {GAI} = \triangle {GAC} $
$ \triangle {GAC}= \frac {1}{3} \triangle {ABC} $
$ GA= \frac {2}{3} AE = \frac {2}{3} q $
similarly $GI = \frac {2}{3} p$ and $AI= \frac {2}{3}r$
So $\triangle {GAI} $ is triangle whose sides are two third of $\triangle {X}$.
So the area of $\triangle {GAI} = \frac {4}{9} \triangle {X}$
$ \frac {4}{9} \triangle {X} = \frac {1}{3} \triangle {ABC}$
$ \frac {4}{3} \triangle {X} = \triangle {ABC} $
