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The image below suggest that that diagonals of a parallelogram are perpendicular exactly when the parallelogram is a rhombus. Use vectors and the dot product to prove this fact.
Let %%vecu%% and %%vecv%% be vectors in %%RR^n%% where %%vecu=(:u_1,u_2,...,u_n:)%% and %%vecv=(:v_1,v_2,...,v_n:)%% and %% ||vecu||=||vecv||%%
Prove that %%(vecu+vecv)%% is perpendicular to %%(vecu-vecv)%%
%%(vecu+vecv) * (vecu-vecv)=(:u_1+v_1,u_2+v_2,...,u_n+v_n:) * (:u_1-v_1,u_2-v_2,...,u_n-v_n:) =%%
Edit: I had assumed two dimensions to start with simply because of the flat picture, but this proof easily extends to n dimensions as well. For the picture, I just copied yours and edited it in paint.
@fritz I like this! How'd you make the picture?
I do think it could be improved slightly, though. Your proof is inherently two dimensional yet the theorem holds in three or even more dimensions as well.
But again - it definitely merits a "Good Answer" vote!