This is write-up 4. Problem #14.

**Given: **Three
medians of a triangle

**Prove: **A) The
medians are concurrent and B) The medians meet at a point whose
distance from any vertex is two thirds the length of the median
from that vertex.

__PROOF:__

__Part A__

Let AM, BN, and CO denote the medians of triangle ABC.

The medians AM and BN intersect at some point Q.

Let point P denote the intersection of ray CQ and side AB.

There is a ray BX in the A-side of BC such that ray BX and ray MA are parallel.

( For a more dynamic sketch that you can move
around click ** here.**
)

In triangle CBD, MQ bisects one side and is
parallel to a second side....**THEREFORE,**
point Q is the midpoint of side CD.

Since NQ bisects two sides of triangle CAD,
NQ is parallel to AD. **THUS**, the
opposite sides of ADBQ are parallel and the figure is a parallelogram.
Its diagonals bisect each other, so, point P is the midpoint of
AB. **THEREFORE** point P and point
O are the same point, which means that the medians are concurrent
at point Q.

__Part B__

If the distance from point N to point Q equals y, then the distance from point A to point D equals 2y which also equals the distance from point Q to point B.

**THEREFORE, **the distance from point B
to point Q equals two thirds the distance from point B to point
N.

If the distance from point P to point Q equals
x, then the distance from point Q to point D equals 2x, and ,
because point Q is the midpoint of CD, the distance from point
Q to point C equals 2x. **THUS**, the
distance from point C to point Q equals two thirds the distance
from point C to point O.

In a similar fashion, it can be shown that the distance from point A to point Q equals two thirds the distance from point A to point M.