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.