The Perpendicular Bisectors of a Triangle are Concurrent

Here we prove that the right bisectors of a triangle are concurrent.


perpendicular-bisector-triangle

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.