I'm kind of confused on how to solve this one, especially through using coordinate proof.
I know how to prove the other way around. If you know that you have a rectangle, then you know that it's diagonals are congruent through use of the distance formula no matter what the coordinates are.
But how would one go about proving the converse? Or at the very least, what theorem am I supposed to be using here? So far, I plotted a rectangle with coordinates $A(0,0), B(c,0), C(c,a), D(0,a).$ Since $\overline{AC} = \overline{BD},$ we have that $\sqrt{c^2+a^2}= \sqrt{(-c)^2+(-a)^2}.$ But where am I supposed to go from here and how does this prove that the parallelogram is a rectangle?
Best Answer
That a rectangle has congruent diagonals was proven above. Going in the other direction, consider the parallelogram $ABCD$ with coordinates $A=(0,0)$, $B=(w,0)$, $C=(w+s,h)$ and $D=(s,h)$, with $w,h>0$. Note that $s$ is the “skew” of the parallelogram. If $AC=BD$, then $$AC^2-BD^2 = ((w+s)^2+h^2)-((w-s)^2+h^2)=4ws=0$$ We know that $w\neq 0$, therefore $s$ must be $0$, i.e., $ABCD$ is a rectangle.