Students are often asked to investigate the medians of triangles, and they are told that the three medians meet at one point. However, a proof of this fact is often omitted, but it is not too difficult to demonstrate this result if a particular corollary is known:

Corollary:If a transversal crosses three or more parallel lines in such a way
as to result in congruent segments between the parallels, then every transversal will
do likewise.

This statement is logically equivalent to a statement that three paralell lines are 'evenly spaced'.

Theorem: The three medians of a triangle are concurrent at a point called the centroid.
Proof. Consider triangle ABC (below) with medians AD and BE.
Construct two points H and G as the midpoints of segments DC and BD, respectively.
It is now necessary to construct appropriate parallel lines so that we may apply the mentioned corollary. Construct four parallel lines 1, l2, l3, and l4 each parallel to line l, the line containing median AD, through points B,G,H, and C, respectivey. The transversal line BC has congruent segments, BG,GD,DG,HC.
Since E is the midpoint of AC, l3 intersects AC at the point E by the above corollary, using AC as the transversal.
In addition, using BE as a transversal, we have that , BF = FR = RE.
Thus the medians AD and BE intersect at R (the centroid!), a point that is two-thirds of the
way from B to E, so BR = (2/3)BE.
Repeating the process using the other two pairs of medians gives the desired concurrent point R, which is the centroid of triangle.