Diagonals of parallelogram intersect at $90^\circ$ if and only if figure is rhombus How can we use vectors and dot products to show that the diagonals of a parallelogram intersect at $90^\circ$ if and only if the figure is a rhombus? 
I did the proof, but I realized my final answer would be a rectangle. (I know a rhombus is a type of rectangle, too). But I only want to prove the two diagonals are orthogonal.
 A: Hint:
The diagonals bisect at $90^\circ \implies m_1m_2=-1$, where $m_1$ and $m_2$ are slopes of the diagonals. That's a parallelogram, opposite sides are equal$ \implies$ magnitude of opposite vectors are equal. Now take into one more property of parallelogram, the diagonals bisect, now you can just use Pythagoras to show sides are equal.
A: write each diagonal (vector) in terms of two adjacent sides (vectors) and show their inner product equals $0$ if and only if the two sides have equal length
A: Suppose you have a parallelogram with vertices A,B,C and D and is oriented as such: AB is parallel to CD (hence AD is parallel to BC). To satisfy that this parallelogram is a rhombus, all sides must be equal. Construct a vector that goes from A to C, $\vec{x_{AC}}$ and another that goes from B to D, $\vec{x_{BD}}$. This shape is a rhombus if and only if $\vec{x_{AC}} \cdot \vec{x_{BD}} = 0$ and the length of the two diagonal vectors are not equal to each other. Another thing you can try to show is that the two diagonals are bisectors of one another, but that is another problem.
(Consider the case where the length of the diagonals are equal, what shape would that be?)
A: Assume the centroid of the parallelogram $P\subset {\mathbb R}^2$ at the origin. Then the four vertices are $a$, $b$, $-a$, $-b$, and the two sidelengths are $|a-b|$ and $|a+b|$. Since
$$|a+b|^2-|a-b|^2=4\ a\cdot b$$
it follows that $P$ is a rhombus iff $a\perp b$.
