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

Explanation

Let `triangle ABC` be a triangle such that `AD` is the median.
Taking `A` as the origin, let the position vectors of `B` and `C` be `vec b`and `vec c`
, respectively.
Then,
Position vector of `D =(vec b+vec c)/2`
Now,
`vec (AD)`=Position vector of D − Position vector of A
`=(vec b+vec c)/2`
`vec(BC)=`Position vector of C − Position vector of B
`=vec c - vec b`
since `vec(AD) bot vec (BC),`
`=>1/2 (vec b+ vec c)*(vec c -vec b)=0`
`=>(vec c+ vec b)* (vec c-vec b)=0`
`=>|vec c|^2-|vec b|^2 =0`
`=> |vec c|=|vec b|`
`=> AC+AB`
Hence `triangle ABC` is an isoscale triangle.