Question

Prove , by vactor methods, that in an isosceles triangle, the median on the base is also the attitude on the base.

Answer 1

We have given isosceles triangle

Let the isosceles $\Delta ABC$.

Let $AB=AC$ and $AD$ is the median then,

$\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})$

$\text{and}\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\left[\text{using}\text{triangle}\text{law}\right]$

Here,

$\overrightarrow{DA}.\overrightarrow{DB}=-\overrightarrow{AD}.\frac{1}{2}\overrightarrow{CB}$

$=-\overrightarrow{AD}.\frac{1}{2}-\overrightarrow{BC}$

$=\frac{1}{2}\overrightarrow{AD}.\overrightarrow{BC}$

$=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC}).\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{AC})$

$=\frac{1}{4}(\overrightarrow{AB}+\overrightarrow{AC}).(\overrightarrow{AC}-\overrightarrow{AB})$

$=\frac{1}{4}[(\overrightarrow{AC}.\overrightarrow{AC})-(\overrightarrow{AB}.\overrightarrow{AB})]$

$=\frac{1}{4}(A{C}^2-A{B}^2)$

$=\frac{1}{4}\times 0$

$=0i.e.\angle ADB={90}^o$

Hence proved by vector method.