Perpendicular Bisectors of a Triangle are Concurrent

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, so

D\left(  {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)


If {m_1} is the slope of AB, then using two point formula to find 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, so its slope m is given as using condition of perpendicular slope is

m  =  - \frac{1}{m} =  - \frac{{{x_2} - {x_1}}}{{{y_2} -  {y_1}}}\,\,\,\,{\text{ -  -  - }}\left( {{\text{ii}}} \right)


Equation of perpendicular OD passing through D with 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 perpendicular bisector OE, we just replace {x_1} by {x_2}, {x_2} by {x_3} and {x_3} by {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 perpendicular bisector OF, we just replace {x_1} by {x_2}, {x_2} by {x_3} and {x_3} by {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.

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