Question

Prove vectorically that median of an isosceles triangle is perpendicular to the base.?

Answer 1

In `DeltaABC`,`AB=AC` and `D` is the mid point of `BC`.

So expressing in vectors we have

`vec(AB)+vec(AC)=2vec(AD)`, since `AD` is half of the diagonal of the parallelogram having adjacent sides `ABandAC`.

So
`vec(AD)=1/2(vec(AB)+vec(AC))`

Now `vec(CB)=vec(AB)-vec(AC)`

So `vec(AD)*vec(CB)`

`=1/2(vec(AB)+vec(AC))*(vec(AB)-vec(AC))`

`=1/2(vec(AB)*vec(AB)- vec(AB)*vec(AC)+ vec(AC)*vec(AB)+ vec(AC)*vec(AC))`

`=1/2(absvec(AB)^2-absvec(AC)^2)`

`=1/2(absvec(AB)^2-absvec(AB)^2)=0`, since `AB=AC`

If `theta` is the angle between `vec(AD)and vec(CB)`

then

`absvec(AD)absvec(CB)costheta=0`

So `theta=90^@`