Medians of a Triangle are Concurrent

The medians of any triangle are concurrent and that the point of concurrency divides each one of them in the ratio 2:1.


medians-triangles

Consider the triangle ABC as shown in the diagram and suppose that A\left( {{x_1},{y_1}} \right), B\left( {{x_2},{y_2}} \right) and C\left( {{x_3},{y_3}} \right) be the vertices of the given triangle ABC. As we know that median is defined as the line segment joining the vertex of triangle to the midpoint to the opposite side of the triangle.

Now suppose that D, E and F are the midpoints of the sides of the triangle ABC, so these midpoints can be calculate using midpoint formula as following:

{\text{Midpoint}}  = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)


Midpoint of the side BC is D = \left( {\frac{{{x_2} +  {x_3}}}{2},\frac{{{y_2} + {y_3}}}{2}} \right)
Midpoint of the side AC is E = \left( {\frac{{{x_1} +  {x_3}}}{2},\frac{{{y_1} + {y_3}}}{2}} \right)
Midpoint of the side AB is F = \left( {\frac{{{x_1} +  {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)
Consider G be the point intersection of the three medians of the triangle ABC.
First we find the coordinates of the point G with respect to median AD, since point G divides the median AD in the ration 2:1, i.e. {k_1}:{k_2}  = 2:1. To find the point G using division formula or ratio formula (internal division).
Let \left(  {{x_1},{y_1}} \right) = \left( {{x_1},{y_1}} \right) and \left( {{x_2},{y_2}} \right) = \left( {\frac{{{x_2}  + {x_3}}}{2},\frac{{{y_2} + {y_3}}}{2}} \right)

\begin{gathered} G = \left( {\frac{{{k_2}{x_1} +  {k_1}{x_2}}}{{{k_1} + {k_2}}},\frac{{{k_2}{y_1} + {k_1}{y_2}}}{{{k_1} +  {k_2}}}} \right) \\ G = \left( {\frac{{\left( 1 \right){x_1} +  \left( 2 \right)\left( {\frac{{{x_2} + {x_3}}}{2}} \right)}}{{1 + 2}},\frac{{\left(  1 \right){y_1} + \left( 2 \right)\left( {\frac{{{y_2} + {y_3}}}{2}}  \right)}}{{1 + 2}}} \right) \\ G = \left( {\frac{{{x_1} + {x_2} +  {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right) \\ \end{gathered}


Second we find the coordinates of the point G with respect to median BE, since point G divides the median BE in the ration 2:1, i.e. {k_1}:{k_2}  = 2:1. To find the point G using division formula or ratio formula (internal division).
Let \left(  {{x_1},{y_1}} \right) = \left( {{x_2},{y_2}} \right) and \left( {{x_2},{y_2}} \right) = \left( {\frac{{{x_1}  + {x_3}}}{2},\frac{{{y_1} + {y_3}}}{2}} \right)

\begin{gathered} G = \left( {\frac{{{k_2}{x_1} + {k_1}{x_2}}}{{{k_1}  + {k_2}}},\frac{{{k_2}{y_1} + {k_1}{y_2}}}{{{k_1} + {k_2}}}} \right) \\ G = \left( {\frac{{\left( 1 \right){x_2} +  \left( 2 \right)\left( {\frac{{{x_1} + {x_3}}}{2}} \right)}}{{1 +  2}},\frac{{\left( 1 \right){y_2} + \left( 2 \right)\left( {\frac{{{y_1} +  {y_3}}}{2}} \right)}}{{1 + 2}}} \right) \\ G = \left( {\frac{{{x_1} + {x_2} +  {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right) \\ \end{gathered}


Similarly, coordinates of G with respect to CF.
Hence, the medians of a triangle are concurrent and its point of concurrency is

G =  \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}}  \right)

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