How to prove the parallelogram law by using geometric method?

Question

How to prove th parallelogram law by using geometric method? NOT ALGEBRA

Answer 1

It can not be fully geometric but you can prove parallelogram law by using law of cosines. If we have paralleogram $ABCD$ then we know that the cosine of angle between $AB$ and $AD$ is equal to minus cosine of angle between $AB$ and $BC$. So by law of cosines $$ BD^2=AD^2+AB^2-2.AD.AB.\cos(DAB)$$ and $$ AC^2=BC^2+AB^2-2.BC.AB.\cos(ABC)$$ Also we know $AD=BC$. Then if we sum up the last two equalities we get law of parallelogram $$BD^2+AC^2=2AB^2+2BC^2.$$

Answer 2

Hint: If the sides of the parallelogram are $a$ and $b$, just evaluate $$\|a+b\|^2+\|a-b\|^2.$$ Very geometrically.

Michael