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 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 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.

Equal-area 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 equal-area 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 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, 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 ma, mb, and mc 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 with sides a, b, c and medians m_a, m_b, m_c,[5]

\tfrac{3}{4}(a+b+c) < m_a+m_b+m_c < a+b+c



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 semi-sum (ma + mb + mc)/2 as σ, we have[7]

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


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.[8]

See also


  1. Weisstein, Eric W. (2010). CRC Concise Encyclopedia of Mathematics, Second Edition. CRC Press. pp. 375–377. ISBN 9781420035223.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
  2. Bottomley, Henry. "Medians and Area Bisectors of a Triangle". Retrieved 27 September 2013.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
  3. Dunn, J. A., and Pretty, J. E., "Halving a triangle," Mathematical Gazette 56, May 1972, 105-108.
  4. Déplanche, Y. (1996). Diccio fórmulas. Medianas de un triángulo. Edunsa. p. 22. ISBN 978-84-7747-119-6. Retrieved 2011-04-24.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
  5. Posamentier, Alfred S., and Salkind, Charles T., Challenging Problems in Geometry, Dover, 1996: pp. 86–87.
  6. Boskoff, Homentcovschi, and Suceava (2009), Mathematical Gazette, Note 93.15.
  7. Benyi, Arpad, "A Heron-type formula for the triangle", Mathematical Gazette 87, July 2003, 324–326.
  8. Leung, Kam-tim; and Suen, Suk-nam; "Vectors, matrices and geometry", Hong Kong University Press, 1994, pp. 53–54

External links