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Question
Show that the perpendicular bisectors of the sides of a triangle are concurrent.
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Solution
Let ABC be a triangle with vertices \[A \left( x_1 , y_1 \right), B \left( x_2 , y_2 \right) \text { and } C \left( x_3 , y_3 \right)\]
Let D, E and F be the midpoints of the sides BC, CA and AB, respectively.
Thus, the coordinates of D, E and F are \[D \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right), E \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right)\] and \[F \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\].
Let
\[m_D , m_E \text { and } m_F\] be the slopes of AD, BE and CF respectively.
\[\therefore\] Slope of BC \[\times\] \[m_D\] = \[-\]1
\[\Rightarrow \frac{y_3 - y_2}{x_3 - x_2} \times m_D = - 1\]
\[ \Rightarrow m_D = - \frac{x_3 - x_2}{y_3 - y_2}\]
Thus, the equation of AD
\[y - \frac{y_2 + y_3}{2} = - \frac{x_3 - x_2}{y_3 - y_2}\left( x - \frac{x_2 + x_3}{2} \right)\]
\[\Rightarrow y - \frac{y_2 + y_3}{2} = - \frac{x_3 - x_2}{y_3 - y_2}\left( x - \frac{x_2 + x_3}{2} \right)\]
\[ \Rightarrow 2y\left( y_3 - y_2 \right) - \left( {y_3}^2 - {y_2}^2 \right) = - 2x\left( x_3 - x_2 \right) + {x_3}^2 - {x_2}^2\]
\[\Rightarrow 2x\left( x_3 - x_2 \right) + 2y\left( y_3 - y_2 \right) - \left( {x_3}^2 - {x_2}^2 \right) - \left( {y_3}^2 - {y_2}^2 \right) = 0\] .......... (1)
Similarly, the respective equations of BE and CF are \[2x\left( x_1 - x_3 \right) + 2y\left( y_1 - y_3 \right) - \left( {x_1}^2 - {x_3}^2 \right) - \left( {y_1}^2 - {y_3}^2 \right) = 0\] ... (2)
\[2x\left( x_2 - x_1 \right) + 2y\left( y_2 - y_1 \right) - \left( {x_2}^2 - {x_1}^2 \right) - \left( {y_2}^2 - {y_1}^2 \right) = 0\] ... (3)
Let
\[L_1 , L_2 \text { and } L_3\]
represent the lines (1), (2) and (3), respectively.
Adding all the three lines,
We observe:
\[1 \cdot L_1 + 1 \cdot L_2 + 1 \cdot L_3 = 0\]
Hence, the perpendicular bisectors of the sides of a triangle are concurrent.
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