Perpendicularity and Dot Product

It's a well-known fact that when two vectors are "perpendicular" (in the Euclidean geometric sense) then their dot products are $0$, i.e. for two vectors $A$ and $B$, $A \cdot B = 0$ iff $A$ and $B$ are perpendicular.

I think one can construct a slightly complicated argument for this fact, starting with the standard basis vectors $e_j \cdot e_k = 0$ for $k \neq j$ and arguing this relation survives a change of basis. You would then have to argue that two perpendicular vectors can then be transformed to a pair of standard basis vectors by the operations of transformation and scaling. Horribly inelegant though - ugh. ("There is no place for in the world for ugly mathematics." - G. H. Hardy) 

Nonetheless, lately, I have been reading Sergei Lang's "Calculus of Several Variables" - highly recommended. In that book, he gives this much more elegant argument.

The argument should be pretty easy to follow for anyone familiar with the dot product. I think You need to see that two triangles are similar in Figure 18.