Question

If the median to the base of a triangle is perpendicular to the base, then triangle is isosceles.

Answer 1

Let ∆ABC be a triangle such that AD is the median. Taking A as the origin, let the position vectors of B and C be $\overrightarrow{b}$ and $\overrightarrow{c}$, respectively. Then,

Position vector of D = $\frac{\overrightarrow{b}+\overrightarrow{c}}{2}$ (Mid-point formula)

Now,

$\overrightarrow{\mathrm{AD}}$ = Position vector of D − Position vector of A = $\frac{\overrightarrow{b}+\overrightarrow{c}}{2}$

$\overrightarrow{\mathrm{BC}}$ = Position vector of C − Position vector of B = $\overrightarrow{c}-\overrightarrow{b}$

Since $\overrightarrow{\mathrm{AD}}\perp \overrightarrow{\mathrm{BC}}$,

$$\begin{gathered} \therefore \overrightarrow{\mathrm{AD}}.\overrightarrow{\mathrm{BC}}=0 \\ \Rightarrow \frac{1}{2}\left(\overrightarrow{b}+\overrightarrow{c}\right).\left(\overrightarrow{c}-\overrightarrow{b}\right)=0 \\ \Rightarrow \left(\overrightarrow{c}+\overrightarrow{b}\right).\left(\overrightarrow{c}-\overrightarrow{b}\right)=0 \\ \Rightarrow {\left|\overrightarrow{c}\right|}^2-{\left|\overrightarrow{b}\right|}^2=0 \\ \Rightarrow \left|\overrightarrow{c}\right|=\left|\overrightarrow{b}\right| \\ \Rightarrow \mathrm{AC}=\mathrm{AB} \end{gathered}$$

Hence, the ∆ABC is an isosceles triangle.