Let D and E are mid point of side AC and side AB respectively. Let us join D and E as shown in left side of figure.
We know that since D and E are mid points two sides of triangle, line DE is parallel to third side BC and DE = (1/2)BC.
Since DE is parallel to BC, as marked in the figure, EDG = GBC and DEG = GCB .
Hence ΔGED and ΔGBC are similar triangles. we have
Hence the intersection point G divides the median BD and CE in the ratio 2:1 or GD = (1/3)BD and GE = (1/3)CE
Now if we draw medians BD and AF as shown in right side of figure, we can similarly prove ΔGFD and ΔGAB are similar,
BD and AF intersects at G so that GF = (1/3)AF and GD = (1/3)BD
since we have proved in both the cases, if any two medians intersect, then intersection point divides the medians in the ratio 1:3,
hence all the three medians are concurrent