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.

In a triangle $ABC$ with medians intersecting at $O$, draw a line throgh $A$ parallel to the median through $B$ and a line through $B$ parallel to the median through $A$. Let $D$ be the intersection of the new lines. Then the parallelogram $AOBD$ has area $2/3$ of the area of the triangle $ABC$ but it can also be partitioned by the diagonal $OD$ into two triangles made out of intervals of lengths $2/3$ of the lengths of the medians themselves. Hence the area of the triangle formed by the medians is $(3/2)^2$ times $1/3$, or $3/4$ of the area of the triangle $ABC$.
