Question

Show that the perpendicular bisectors of the sides of a triangle are concurrent.

Answer 1

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.