Question

Show that the median to the base of an isosceles triangle is perpendicular to base.

Answer 1

Let $ABC$ be an isosceles triangle in which $|\overrightarrow{AB}|=|\overrightarrow{AC}|$
Let $A$ be the point of reference (origin) and let
$\overrightarrow{AB}=\overrightarrow{b},\overrightarrow{AC}=\overrightarrow{c}$
Let $D$ be the middle point $BC$.
Then, $AD=\frac{\overrightarrow{b}+\overrightarrow{c}}{2}$
Now, $BC=\overrightarrow{c}-\overrightarrow{b}$
$\therefore \overrightarrow{AD}\cdot \overrightarrow{BC}=\left(\frac{\overrightarrow{b}+\overrightarrow{c}}{2}\right)\cdot (\overrightarrow{c}-\overrightarrow{b})=\frac{1}{2}(|\overrightarrow{c}{|}^2-|\overrightarrow{b}{|}^2)=0$ [ Since $|\overrightarrow{b}|=|\overrightarrow{c}|$

So $\overrightarrow{AD}.\overrightarrow{BC}=0\Rightarrow$ $\overrightarrow{AD}$ is perpendicular to $\overrightarrow{BC}$.

Hence the problem,