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 from each vertex, and they all intersect each other at the triangle's centroid. In the case of isosceles and equilateral triangles, a median bisects any angle at a vertex whose two adjacent sides are equal in length.
The concept of a median extends to tetrahedra.
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 the intersection point of the medians.
Equalarea division
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.
Proof of equalarea property
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 (most commonly denoted G).
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 onehalf 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, one 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 m_{a}, m_{b}, and m_{c} from their midpoints.
Thus we have the relationships:^{[4]}

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,^{[5]}

\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,^{[5]}

\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.^{[6]}
The medians of a right triangle with hypotenuse c satisfy m_a^2+m_b^2=5m_c^2.
Any triangle's area T can be expressed in terms of its medians m_a, m_b, and m_c as follows. Denoting their semisum (m_{a} + m_{b} + m_{c})/2 as σ, we have^{[7]}

T = \frac{4}{3} \sqrt{\sigma (\sigma  m_a)(\sigma  m_b)(\sigma  m_c)}.
Tetrahedron
A tetrahedron is a threedimensional 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.^{[8]}
See also
References

^ Weisstein, Eric W. (2010). CRC Concise Encyclopedia of Mathematics, Second Edition. CRC Press. pp. 375–377.

^ Bottomley, Henry. "Medians and Area Bisectors of a Triangle". Retrieved 27 September 2013.

^ Dunn, J. A., and Pretty, J. E., "Halving a triangle," Mathematical Gazette 56, May 1972, 105108.

^ Déplanche, Y. (1996). Diccio fórmulas. Medianas de un triángulo. Edunsa. p. 22.

^ ^{}a ^{b} Posamentier, Alfred S., and Salkind, Charles T., Challenging Problems in Geometry, Dover, 1996: pp. 86–87.

^ Boskoff, Homentcovschi, and Suceava (2009), Mathematical Gazette, Note 93.15.

^ Benyi, Arpad, "A Herontype formula for the triangle", Mathematical Gazette 87, July 2003, 324–326.

^ Leung, Kamtim; and Suen, Suknam; "Vectors, matrices and geometry", Hong Kong University Press, 1994, pp. 53–54
External links

Medians and Area Bisectors of a Triangle

The Medians at cuttheknot

Area of Median Triangle at cuttheknot

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.
This article was sourced from Creative Commons AttributionShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, EGovernment Act of 2002.
Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.
By using this site, you agree to the Terms of Use and Privacy Policy. World Heritage Encyclopedia™ is a registered trademark of the World Public Library Association, a nonprofit organization.