# The 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. 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 $C\left( {{x_3},{y_3}} \right)$ are the vertices of the given triangle $ABC$. As we know, the median is defined as the line segment joining the vertex of the 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 calculated using the midpoint formula as follows:
${\text{Midpoint}} = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)$

The midpoint of side $BC$ is $D = \left( {\frac{{{x_2} + {x_3}}}{2},\frac{{{y_2} + {y_3}}}{2}} \right)$

The midpoint of side $AC$ is $E = \left( {\frac{{{x_1} + {x_3}}}{2},\frac{{{y_1} + {y_3}}}{2}} \right)$

The midpoint of side $AB$ is $F = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)$

Consider $G$ is  the point intersection of the three medians of the triangle $ABC$.

First we find the coordinates of the point $G$ with respect to the median $AD$, since the point $G$ divides the median $AD$ in the ratio $2:1$, i.e. ${k_1}:{k_2} = 2:1$. We find the point $G$ using the 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 the median $BE$, since the point $G$ divides the median $BE$ in the ratio $2:1$, i.e. ${k_1}:{k_2} = 2:1$. We find the point $G$ using the 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, the coordinates of $G$ with respect to $CF$.

Hence, the medians of the triangle are concurrent and the point of concurrency is
$G = \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$