# Median (geometry)

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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. In the case of isosceles and equilateral triangles, a median bisects any angle at a vertex whose two adjacent sides are equal in length.

## Relation to center of mass

Each median of a triangle passes through the triangle's centroid, which is the center of mass of an object of uniform density in the shape of the triangle.[1] Thus the object would balance on any line through the centroid, including any median.

## Equal-area division

Each median divides the area of the triangle in half; hence the name. (Any other lines which divide the area of the triangle into two equal parts do not pass through the centroid.)[2] The three medians divide the triangle into six smaller triangles of equal area.

### 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 $[ADO]=[BDO], [AFO]=[CFO], [BEO]=[CEO],$ and $[ABE]=[ACE] \,$, where $[ABC]$ represents the area of triangle $\triangle ABC$ ; these hold because in each case the two triangles have bases of equal length and share a common altitude from the (extended) base, and a triangle's area equals one-half its base times its height.

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 $[AFO]=[FCO]=[DBO]=[ADO]=[BEO]=[CEO] \,$.

## Formulas involving the medians' lengths

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

$m_a = \sqrt {\frac{2 b^2 + 2 c^2 - a^2}{4} },$
$m_b = \sqrt {\frac{2 a^2 + 2 c^2 - b^2}{4} },$
$m_c = \sqrt {\frac{2 a^2 + 2 b^2 - c^2}{4} },$

where a, b and c are the sides of the triangle with respective medians ma, mb, and mc from their midpoints.

Thus we have the relationships:[3]

$a = \frac{2}{3} \sqrt{-m_a^2 + 2m_b^2 + 2m_c^2} = \sqrt{2(b^2+c^2)-4m_a^2} = \sqrt{\frac{b^2}{2} - c^2 + 2m_b^2} = \sqrt{\frac{c^2}{2} - b^2 + 2m_c^2},$
$b = \frac{2}{3} \sqrt{-m_b^2 + 2m_a^2 + 2m_c^2} = \sqrt{2(a^2+c^2)-4m_b^2} = \sqrt{\frac{a^2}{2} - c^2 + 2m_a^2} = \sqrt{\frac{c^2}{2} - a^2 + 2m_c^2},$
$c = \frac{2}{3} \sqrt{-m_c^2 + 2m_b^2 + 2m_a^2} = \sqrt{2(b^2+a^2)-4m_c^2} = \sqrt{\frac{b^2}{2} - a^2 + 2m_b^2} = \sqrt{\frac{a^2}{2} - b^2 + 2m_a^2}.$

## Other properties

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,[4]

$\tfrac{3}{4}$(perimeter) < sum of the medians < (perimeter of the triangle).

For any triangle with sides $a, b, c$ and medians $m_a, m_b, m_c$,[4]

$\tfrac{3}{4}(a^2+b^2+c^2)=m_a^2+m_b^2+m_c^2.$

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

The medians of a right triangle with hypotenuse c satisfy $m_a^2+m_b^2=5m_c^2.$

We can express any triangle's area T in terms of its medians ma, mb, and mc as follows. Denoting their semi-sum (ma + mb + mc)/2 as σ, we have[6]

$T = \frac{4}{3} \sqrt{\sigma (\sigma - m_a)(\sigma - m_b)(\sigma - m_c)}.$

## References

1. ^ Weisstein, Eric W. (2010). CRC Concise Encyclopedia of Mathematics, Second Edition. CRC Press. pp. 375–377. ISBN 9781420035223.
2. ^ Bottomley, Henry. "Medians and Area Bisectors of a Triangle". Retrieved 27 September 2013.
3. ^ Déplanche, Y. (1996). Diccio fórmulas. Medianas de un triángulo. Edunsa. p. 22. ISBN 978-84-7747-119-6. Retrieved 2011-04-24.
4. ^ a b Posamentier, Alfred S., and Salkind, Charles T., Challenging Problems in Geometry, Dover, 1996: pp. 86-87.
5. ^ Boskoff, Homentcovschi, and Suceava (2009), Mathematical Gazette, Note 93.15.
6. ^ Benyi, Arpad, "A Heron-type formula for the triangle," Mathematical Gazette" 87, July 2003, 324–326.