Median (geometry)

Each median of a triangle passes through the triangle's centroid, which is the center of mass of an infinitely thin object of uniform density coinciding with the triangle.[1] Thus the object would balance on the intersection point of the medians. The centroid is twice as close along any median to the side that the median intersects as it is to the vertex it emanates from.

Each median divides the area of the triangle in half; hence the name, and hence a triangular object of uniform density would balance on any median. (Any other lines which divide the area of the triangle into two equal parts do not pass through the centroid.)[2][3] The three medians divide the triangle into six smaller triangles of equal area.

The lengths of the medians can be obtained from Apollonius' theorem as:

${\displaystyle m_{a}={\sqrt {\frac {2b^{2}+2c^{2}-a^{2}}{4}}},}$

${\displaystyle m_{b}={\sqrt {\frac {2a^{2}+2c^{2}-b^{2}}{4}}},}$

${\displaystyle m_{c}={\sqrt {\frac {2a^{2}+2b^{2}-c^{2}}{4}}},}$

where a, b and c are the sides of the triangle with respective medians m_a, m_b, and m_c from their midpoints.

Thus we have the relationships:[4]

${\displaystyle a={\frac {2}{3}}{\sqrt {-m_{a}^{2}+2m_{b}^{2}+2m_{c}^{2}}}={\sqrt {2(b^{2}+c^{2})-4m_{a}^{2}}}={\sqrt {{\frac {b^{2}}{2}}-c^{2}+2m_{b}^{2}}}={\sqrt {{\frac {c^{2}}{2}}-b^{2}+2m_{c}^{2}}},}$

${\displaystyle b={\frac {2}{3}}{\sqrt {-m_{b}^{2}+2m_{a}^{2}+2m_{c}^{2}}}={\sqrt {2(a^{2}+c^{2})-4m_{b}^{2}}}={\sqrt {{\frac {a^{2}}{2}}-c^{2}+2m_{a}^{2}}}={\sqrt {{\frac {c^{2}}{2}}-a^{2}+2m_{c}^{2}}},}$

${\displaystyle c={\frac {2}{3}}{\sqrt {-m_{c}^{2}+2m_{b}^{2}+2m_{a}^{2}}}={\sqrt {2(b^{2}+a^{2})-4m_{c}^{2}}}={\sqrt {{\frac {b^{2}}{2}}-a^{2}+2m_{b}^{2}}}={\sqrt {{\frac {a^{2}}{2}}-b^{2}+2m_{a}^{2}}}.}$

Let ABC be a triangle, let G be its centroid, and let D, E, and F be the midpoints of BC, CA, and AB, respectively. For any point P in the plane of ABC then

${\displaystyle PA+PB+PC\leq 2(PD+PE+PF)+3PG.}$[5]

The centroid divides each median into parts in the ratio 2:1, with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex.

For any triangle with sides ${\displaystyle a,b,c}$ and medians ${\displaystyle m_{a},m_{b},m_{c},}$[6]

${\displaystyle {\tfrac {3}{4}}(a+b+c)<m_{a}+m_{b}+m_{c}<a+b+c}$

and

${\displaystyle {\tfrac {3}{4}}(a^{2}+b^{2}+c^{2})=m_{a}^{2}+m_{b}^{2}+m_{c}^{2}.}$

The medians from sides of lengths a and b are perpendicular if and only if ${\displaystyle a^{2}+b^{2}=5c^{2}.}$[7]

The medians of a right triangle with hypotenuse c satisfy ${\displaystyle m_{a}^{2}+m_{b}^{2}=5m_{c}^{2}.}$

Any triangle's area T can be expressed in terms of its medians ${\displaystyle m_{a},m_{b}}$, and ${\displaystyle m_{c}}$ as follows. Denoting their semi-sum (m_a + m_b + m_c)/2 as σ, we have[8]

${\displaystyle T={\frac {4}{3}}{\sqrt {\sigma (\sigma -m_{a})(\sigma -m_{b})(\sigma -m_{c})}}.}$

medians of a tetrahedron

${\displaystyle {\begin{aligned}&{\frac {|AS|}{|SS_{BCD}|}}={\frac {|BS|}{|SS_{ACD}|}}={\frac {|CS|}{|SS_{ABD}|}}\\=&{\frac {|DS|}{|SS_{ABC}|}}={\frac {3}{1}}\end{aligned}}}$

A tetrahedron is a three-dimensional object having four triangular faces. A line segment joining a vertex of a tetrahedron with the centroid of the opposite face is called a median of the tetrahedron. There are four medians, and they are all concurrent at the centroid of the tetrahedron.[9] As in the two-dimensional case, the centroid of the tetrahedron is the center of mass. However contrary to the two-dimensional case the centroid divides the medians not in a 2:1 ratio but in a 3:1 ratio (Commandino's theorem).

• Angle bisector
• Altitude (triangle)
• Automedian triangle

Wikimedia Commons has media related to Median (geometry).

• 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
• Weisstein, Eric W. "Triangle Median". MathWorld.