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)%%

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.

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!

## Comments

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:) =%%

%%(u_1+v_1)(u_1-v_1)+(u_2+v_2)(u_2-v_2)+...+(u_n+v_n)(u_n-v_n)=u_1^2-v_1^2+u_2^2-v_2^2+...+u_n^2-v_n^2=%%

%%(u_1^2+u_2^2+...+u_n^2)-(v_1^2+v_2^2+...+v_n^2)=||vecu||^2-||vecv||^2%%

%%=||vecu||^2-||vecu||^2=0%%

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!