Let `vec(a)` and `vec(b)` be two nonzero vector. Prove that
`vec(a) bot vec(b) hArr |vec(a)+vec(b)|=|vec(a)-vec(b)|`.

Explanation

Let `vec(a) bot vec(b)`. Then, `(vec(a).vec(b))=0` ...(i)
Now, `|vec(a)+vec(b)|^(2)=(vec(a)+vec(b)).(vec(a)+vec(b))`
`=vec(a).vec(a)+vec(a).vec(b)+vec(b).vec(a)+vec(b).vec(b)`
`=|vec(a)|^(2)+|vec(b)|^(2) { :' vec(b).vec(b)=0 and vec(b).vec(a)=vec(a).vec(b)=0}`.
Also, `|vec(a)-vec(b)|^(2)=(vec(a)-vec(b)).(vec(a)-vec(b))`
`=vec(a).vec(a)-vec(a).vec(b)-vec(b).vec(a)+vec(b).vec(b)`
`= |vec(a)|^(2)+|vec(b)|^(2) { :' vec(a).vec(b)=0 and vec(b).vec(a)=vec(a).vec(b)=0}`
Thus, `|vec(a)+vec(b)|^(2)=|vec(a)-vec(b)|^(2)`, and therefore, `|vec(a)+vec(b)|=|vec(a)-vec(b)|`.
`:. vec(a) bot vec(b) implies |vec(a)+vec(b)|=|vec(a)-vec(b)|`.
Conversely, suppose that `|vec(a)+vec(b)|=|vec(a)-vec(b)|`. Then,
`|vec(a)+vec(b)|=|vec(a)-vec(b)|implies |vec(a)+vec(b)|^(2)=|vec(a)-vec(b)|^(2)`
`implies (vec(a)+vec(b)).(vec(a)+vec(b))=(vec(a)-vec(b)).(vec(a)-vec(b))`
`implies vec(a).vec(a)+vec(a).vec(b)+vec(b).vec(a)+vec(b).vec(b)`
`=vec(a).vec(a)-vec(a).vec(b)-vec(b).vec(a)+vec(b).vec(b)`
`implies 2(vec(a).vec(b)+vec(b).vec(a))=0`
`implies 4(vec(a).vec(b))=0" "[ :' vec(b).vec(a)=vec(a).vec(b)]`
`implies vec(a).vec(b)=0implies vec(a) bot vec(b)`.
Thus, `|vec(a)+vec(b)|=|vec(a)-vec(b)|implies vec(a) bot vec(b)`.
Hence, `|vec(a)+vec(b)|=|vec(a)-vec(b)| hArr vec(a) bot vec(b)`.