# The Perpendicular Bisectors of a Triangle are Concurrent

Here we prove that the right bisectors 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 $ABC$. Let $D$, $E$ and $F$ be the midpoints of $AB$, $BC$ and $CA$ respectively. Since $D$ is the midpoint of $AB$, then
$D\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)$

If ${m_1}$ is the slope of $AB$, then we use the two point formula to find the slope of line
${m_1} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Since the perpendicular bisector $OD$ is perpendicular to the side $AB$, its slope $m$ is given by using the condition of a perpendicular slope:
$m = – \frac{1}{m} = – \frac{{{x_2} – {x_1}}}{{{y_2} – {y_1}}}\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

The equation of perpendicular $OD$ passing through $D$ with the slope $m$ is
$\begin{gathered} y – {y_3} = m\left( {x – {x_3}} \right) \\ \Rightarrow y – \frac{{{y_1} + {y_2}}}{2} = – \frac{{{x_2} – {x_1}}}{{{y_2} – {y_1}}}\left( {x – \frac{{{x_1} + {x_2}}}{2}} \right) \\ \Rightarrow \left( {{y_2} – {y_1}} \right)y – \left( {{y_2} – {y_1}} \right)\left( {\frac{{{y_1} + {y_2}}}{2}} \right) = – \left( {{x_2} – {x_1}} \right)x + \left( {{x_2} – {x_1}} \right)\left( {\frac{{{x_1} + {x_2}}}{2}} \right) \\ \Rightarrow \left( {{x_2} – {x_1}} \right)x + \left( {{y_2} – {y_1}} \right)y – \frac{{\left( {{x_2} – {x_1}} \right)\left( {{x_1} + {x_2}} \right)}}{2} – \frac{{\left( {{y_2} – {y_1}} \right)\left( {{y_1} + {y_2}} \right)}}{2} = 0 \\ \Rightarrow \left( {{x_2} – {x_1}} \right)x + \left( {{y_2} – {y_1}} \right)y – \frac{1}{2}\left( {{x_2}^2 – {x_1}^2} \right) – \frac{1}{2}\left( {{y_2}^2 – {y_1}^2} \right) = 0\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right) \\ \end{gathered}$

For the equation of the perpendicular bisector $OE$, 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
$\Rightarrow \left( {{x_3} – {x_2}} \right)x + \left( {{y_3} – {y_2}} \right)y – \frac{1}{2}\left( {{x_3}^2 – {x_2}^2} \right) – \frac{1}{2}\left( {{y_3}^2 – {y_2}^2} \right) = 0\,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right)$

For the equation of the perpendicular bisector $OF$, 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
$\Rightarrow \left( {{x_1} – {x_3}} \right)x + \left( {{y_1} – {y_3}} \right)y – \frac{1}{2}\left( {{x_1}^2 – {x_3}^2} \right) – \frac{1}{2}\left( {{y_1}^2 – {y_3}^2} \right) = 0\,\,\,{\text{ – – – }}\left( {\text{v}} \right)$

To see whether the perpendicular bisector (iii), (iv) and (v) are concurrent, consider the determinant:
$\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{ – \frac{1}{2}\left( {{x_2}^2 – {x_1}^2} \right) – \frac{1}{2}\left( {{y_2}^2 – {y_1}^2} \right)} \\ {{x_3} – {x_2}}&{{y_3} – {y_2}}&{ – \frac{1}{2}\left( {{x_3}^2 – {x_2}^2} \right) – \frac{1}{2}\left( {{y_3}^2 – {y_2}^2} \right)} \\ {{x_1} – {x_3}}&{{y_1} – {y_3}}&{ – \frac{1}{2}\left( {{x_1}^2 – {x_3}^2} \right) – \frac{1}{2}\left( {{y_1}^2 – {y_3}^2} \right)} \end{array}} \right|$
${R_3} + \left( {{R_1} + {R_2}} \right)$
$= \left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{ – \frac{1}{2}\left( {{x_2}^2 – {x_1}^2} \right) – \frac{1}{2}\left( {{y_2}^2 – {y_1}^2} \right)} \\ {{x_3} – {x_2}}&{{y_3} – {y_2}}&{ – \frac{1}{2}\left( {{x_3}^2 – {x_2}^2} \right) – \frac{1}{2}\left( {{y_3}^2 – {y_2}^2} \right)} \\ 0&0&0 \end{array}} \right| = 0$

This shows that the perpendicular bisectors of the triangle are concurrent.