# 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 $C\left( {{x_3},{y_3}} \right)$ 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:

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 $\left( {{x_2},{y_2}} \right) = \left( {\frac{{{x_2} + {x_3}}}{2},\frac{{{y_2} + {y_3}}}{2}} \right)$

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 $\left( {{x_2},{y_2}} \right) = \left( {\frac{{{x_1} + {x_3}}}{2},\frac{{{y_1} + {y_3}}}{2}} \right)$

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