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*, l*2*, l*3*, *and *l*4 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*, *l*3 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.