common point of triangle medians

Theorem.  The three medians ( of a triangle intersect one another in one point, which divides each median in the ratio 2:1.


Proof.  Let the medians of a triangle  ABC  be AD, BE and CF.  Any median vector is the arithmetic meanMathworldPlanetmath of the side vectors emanating from the same vertex.  Using vectors, let us form three ways all beginning from the vertex A, the first going simply 2/3 of the median vector AD ( in the picture):

23AD=2312(AB+AC)=13(AB+AC) (1)

The second way goes first the side vector AB and then 2/3 of the median vector BE (green in the picture):

AB+23BE=AB+2312[-AB+(AC-AB)]=13(AB+AC) (2)

Similarly, the third way goes first the side vector AC and then 2/3 of the median vector CF (red in the picture):

AC+23CF=AC+2312[-AC+(AB-AC)]=13(AB+AC) (3)

Thus the ways (2) and (3), where one goes from A to another vertex and continues along the corresponding median 2/3 of its length, lead to the point M which is attained directly along AD.  This means that all medians intersect in M.  The distanceMathworldPlanetmath of M from any vertex is 2/3 of the corresponding median, and so the rest of the median is 1/3 of its length, i.e. the ratio of the parts of any median is 2:1.

Title common point of triangle medians
Canonical name CommonPointOfTriangleMedians
Date of creation 2013-03-22 17:46:54
Last modified on 2013-03-22 17:46:54
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 8
Author pahio (2872)
Entry type Theorem
Classification msc 51M04
Related topic MutualPositionsOfVectors
Related topic ParallelogramPrinciple
Related topic DifferenceOfVectors
Related topic TriangleMidSegmentTheorem
Related topic LengthsOfTriangleMedians