Prove that the median of a triangle divides it into two triangles, the areas of which are equal to each other.

Let the segment BM be the median drawn to the AC side of the triangle ABC, then the segment AM = CM, since the median is we divide the AC side in half.

From the vertex B we construct the height ВН, which is both the height of the ABM triangle and the BCM triangle.

Then Svm = AM * ВН / 2, Svcm = CM * ВН/ 2, and since CM = AM, then Svcm = AM * ВН / 2 = Savm.

Sawm = Svcm, as required.