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.
Best Answer
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.