So, I've dealt with the parallelogram law in various ways a bunch of times now. And algebraically (or maybe I mean arithmetically) it makes perfect sense to me--I can prove it, understand it, and I accept it. However, most authors make an appeal to a picture of a parallelogram with diagonals or vectors draw in, like this one from Wikipedia:
This sort of makes sense to me but at the end of the day, the parallelogram law is relating squares of these line segments not the segments themselves (directly), so I don't really gain any intuition from this diagram. Can somebody spread some geometric light on this without appealing to algebra (or with minimal appeal)? If it's the standard proof (i.e. similar to the Wikipedia proof), I'm not interested. Alternatively, if someone can show why my request isn't possible that would cool too.
Thank you.
We can use also the Pythagoras's theorem.
Let $AE$ and $BF$ be altitudes of $ABCD$, $E\in DC$, $C\in EF$, $AD=a$, $AB=b$ and $EC=c$.
Now, since $\Delta ADE\cong\Delta BCF$, $$DE=CF=b-c,$$ $AE=BF$, $$DF=DC+CF=b+b-c=2b-c$$ and by the Pythagoras's theorem we obtain: $$AC^2+BD^2=(AE^2+EC^2)+(DF^2+BF^2)=2AE^2+EC^2+DF^2=$$ $$=2(a^2-(b-c)^2))+c^2+(2b-c)^2=2(a^2+b^2)=AB^2+BC^2+CD^2+DA^2$$ and we done!