# Concurrency of Triangle Medians

# By

Kenneth E. Montgomery

Given
(Figure
1), we construct midpoints of the three segments and have
,
and
.

**Figure 1:**

The median of a triangle is defined
to be a segment whose endpoints are a vertex of a triangle and the midpoint of
the opposite side.

**Figure 2:
**,
with medians:
,
and

By constructing segments connecting
each vertex to the midpoint of the opposite side, we have medians:
,
and
(Figure
2). Open the file KM04Tri.gsp to construct the
midpoints and medians (Figure 2). It appears that the three medians share a
mutual intersection at *P*. Open the file KM04Anim.gsp
for an animation of
which demonstrates the properties of *P*, as the
size and shape of
is changed. For further exploration, open the file Assign04KM.gsp to observe *P* for your own
configuration of
or simply click on the intersection of the three
medians in your construction, from above and label the point *P*.

## Concurrency

If three or more lines intersect at
a common point, the lines are said to be *concurrent*. The point of
intersection for such lines is said to be a *point of concurrency*. P is
the point of concurrency for the medians of a triangle, which is known as the *centroid*

*Lemma**: Parallel
lines divide all transversals in the in the same proportion.*

**Proof of Lemma:** Construct
two lines *m* and *n*, such that
(Fig. 3).

**Figure 3**: Parallel Lines, *m*
and *n*

Let *m* and *n* be cut by
two transversals, *p* and *q*, such that *p* and *q* are
not parallel. Since p and q are not parallel, the two lines intersect at point
A, forming
.

**Figure 4**: Lines *m* and *n* cut by
transversals *p* and *q*, intersecting at A

The lines *p* and *q*
intersect *m* and *n* at points C, E, B and D.
and
since
and corresponding angles are congruent, and
, by the reflexive property. Therefore,
by AAA similarity. Corresponding sides of similar
triangles are proportional, so
and
, by same proportion. Thus, parallel lines divide all
transversals in the same proportion.Ò

**Theorem:**

The medians of a triangle intersect
at a point (the *centroid*, *P*) that is two thirds of the distance
from each vertex to the midpoint of the opposite side.

**Proof:**

**Figure 3:
**, with medians:
,
and
; and constructed segments
and

Beginning with triangle
, construct midpoints
,
and
. Construct medians,
,
and
. Construct midpoints

and
. Construct lines segments, parallel to
, through points G and H. By similar triangles, we
have *FB* = *AF*, so *BG* = *GD* and *EC* = *AC*,
so *HC* = *DH*. Therefore,
, since *D* = M(
). Take
to be a transversal of the three parallel lines,
,
and
. By the lemma,
parallel lines divide all transversals in the same proportion, so we see that
. Thus
is divided into 3 equal points and *P* lies
two-thirds of the distance from *B* to *E*. Likewise,
and *P* is two-thirds the distance from *C*
to *F*, so
is concurrent with
at point *P*. Without loss of generality, *P*
is two-thirds the distance from *A* to *D*. and the three medians are
concurrent at *P*.

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