Median: Information from Answers.com

For another use of the term median in geometry, see Geometric median.

The triangle medians and the centroid.

In geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposing side. Every triangle has exactly three medians; one running from each vertex to the opposite side. The median only bisects the vertex angle from which it is drawn in the case of equilateral triangles.

1 Point of concurrency
2 Equal-area division
- 2.1 Proof
3 Formula for length
4 See also
5 External links

Point of concurrency

As a corollary of Ceva's theorem, the three medians are concurrent. The point of concurrency is known as the triangle's centroid, or center of mass of the triangle. Note that this means that the centroid is always in the interior of the triangle. Two-thirds of the length of each median is between the vertex and the centroid, while one-third is between the centroid and the midpoint of the opposite side.

Equal-area division

Each median divides the triangle in half; hence the name. The three medians divide the triangle into six smaller triangles of equal area.

Any other lines which divide the area of the triangle into two equal parts do not pass through the centroid.

Proof

Consider a triangle ABC. Let D be the midpoint of $\overline{AB}$ , E be the midpoint of $\overline{BC}$ , F be the midpoint of $\overline{AC}$ , and O be the centroid.

By definition, $AD=DB, AF=FC, BE=EC \,$ , thus $[ADO]=[BDO], [AFO]=[CFO], [BEO]=[CEO], [ABE]=[ACE] \,$ , where $[A B C]$ represents the area of triangle $\triangle ABC$ .

We have:

$[ABO]=[ABE]-[BEO] \,$

$[ACO]=[ACE]-[CEO] \,$

Thus, $[ABO]=[ACO] \,$ and $[ADO]=[DBO], [ADO]=\frac{1}{2}[ABO]$

Since $[AFO]=[FCO], [AFO]= \frac{1}{2}ACO=\frac{1}{2}[ABO]=[ADO]$ , therefore, $[AFO]=[FCO]=[DBO]=[ADO]\,$ . Using the same method, you can show that $[AFO]=[FCO]=[DBO]=[ADO]=[BEO]=[CEO] \,$ .