# The Altitudes of a Triangle are Concurrent

Here we prove that the altitudes of a triangle are concurrent. Let $A\left( {{x_1},{y_1}} \right)$, $B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$ be the vertices of the triangle $ABC$. If ${m_1}$ is the slope of $AB$, then we use the two point formula to find the slope of the line

Since the altitude $CD$ is perpendicular to the side $AB$, its slope $m$ is given by using a condition of the perpendicular slope:

The equation of altitude $CD$ passing through $C\left( {{x_3},{y_3}} \right)$ with slope $m$ is

For the equation of altitude $AB$, we just replace ${x_1}$ with ${x_2}$, ${x_2}$ with ${x_3}$ and ${x_3}$ with ${x_1}$ in (iii) (i.e. ${x_1} \to {x_2},\,\,{x_2} \to {x_3},\,\,{x_3} \to {x_1}$), so

For the equation of altitude $BF$, we just replace ${x_1}$ with ${x_2}$, ${x_2}$ with ${x_3}$ and ${x_3}$ with ${x_1}$ in (iv) (i.e. ${x_1} \to {x_2},\,\,{x_2} \to {x_3},\,\,{x_3} \to {x_1}$), so

To see whether altitudes (iii), (iv) and (v) are concurrent, consider the determinant:

This shows that the altitudes of the triangle are concurrent.