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.


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) 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)