  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. thus the object would balance on the intersection point of the medians. ==equal-area division== thumb|right|300px| 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.)dunn, j. a., and pretty, j. e., "halving a triangle," mathematical gazette 56, may 1972, 105-108. 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 med
Elementary_geometry Triangles