In a triangle ABC the points D, E and F are the midpoints of AB, BC and CA respectively.
The medians intersect at the point O inside the triangle. Drawing a picture will help you to follow the argument that follows.
Let Z(XYZ) mean the area of triangle XYZ, where X, Y and Z are the representative vertices of a triangle XYZ.
Z(COE)=Z(EOB)=x because the triangles have the same height and the same length base because CE=EB.
Similarly Z(AOF)=Z(FOC)=z and Z(BOD)=Z(DOA)=y. The values of x, y and z are defined here as a shorthand convenience. Write x, y and z in the relevant triangles and you will see that the following equations are true. Z(CAE)=2z+x=Z(EAB)=2y+x, Z(ABF)=2y+z=Z(FBC)=2x+z and Z(BCD)=2x+y=Z(DCA)=2z+y. The area of ABC is 2(x+y+z). From these equalities, 2z+x=2y+x, therefore z=y; 2y+z=2x+z, so x=y. Therefore x=y=z and the area of ABC=2(3x)=6x. In other words, the area of ABC is 6 times the area of one of the equal-area triangular partitions.