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.

Point of concurrency

The three medians are concurrent at a point 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

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

Formula for length

Applying Stewart's theorem one gets:

$m\; =\; sqrt\; \{frac\{2\; b^2\; +\; 2\; c^2\; -\; a^2\}\{4\}\; \}$

where a is the side of the triangle whose midpoint is the extreme point of median m.

See also

• Angle bisector
• Altitude (triangle)
• Ceva's theorem

External links

• Medians and Area Bisectors of a Triangle
• The Medians at cut-the-knot
• Area of Median Triangle at cut-the-knot
• Medians of a triangle With interactive animation
• Constructing a median of a triangle with compass and straightedge animated demonstration