Question

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

Answer 1

Let ABC be a triangle with vertices A $(x_1,y_1),B(x_2,y_2)$ and C $(x_3,y_3)$ 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

$D(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$

$E(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})$ and

$F(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Let $m_D,m_E$ 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

$y-\frac{y_2-y_3}{2}=-\frac{x_3-x_2}{y_3-y_2}(x-\frac{x_2+x_3}{2})$

$\Rightarrow y=\frac{y_2-y_3}{2}=-\frac{x_3-x_2}{y_3-y_2}(x-\frac{x_2+x_3}{2})$

$\Rightarrow 2y(y_3-y_2)-(y_3^2-y_2^2)=2x(x_3-x_2)+x_3^2-x_2^2$

$\Rightarrow 2x(x_3-x_2)+2y(y_3-y_2)-(x_3^2-x_2^2)-(y_3^2-y_2^2)=0...\left(i\right)$

Similarly, the respective equations of BE and CF are

$2x(x_1-x_3)+2y(y_1-y_3)-(x_1^2-x_3^2)-(y_1^2-y_3^2)=0...\left(ii\right)$

$2x(x_2-x_1)+2y(y_2-y_1)-(x_2^2-x_1^2)-(y_2^2-y_1^2)=0...\left(iii\right)$

Let $L_1,L_2$ and $L_3$ respresent the lines (i), (ii) and (iii), respectively.

Adding all the three lines.

We observe:

$1l{L}_1+1.L_2+1.L_3=0$

Hence, the perpendicular bisectors of the sides of a triangle are concurrent.