# Give a coordinate proof that a parallelogram is a rectangle iff its diagonals are congruent

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?

• Hint: compute $AC-BD$. You’ll have an easier time of it if you look at the squares of the lengths instead. – amd Sep 11 '15 at 18:07
• AC - BD = 0. Is that the end of my proof? I wanted to use the triangle made up by the diagonals to prove that the parallelogram had 4 right angles. But I'm not sure if I'd be answering the question at that point. – Alphatron Sep 12 '15 at 4:02
• That’s where your proof starts—you’re working on the congruent diagonals => rectangle direction. Use the coordinate-based expressions for these lengths and see where that takes you. – amd Sep 13 '15 at 4:35

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.
Suppose you have a parallelogram whose sides are determined by the plane position vectors $$a$$ and $$b,$$ then its diagonals are $$a+b$$ and $$a-b$$ respectively. That the diagonals are congruent implies that $$|a+b|=|a-b|.$$ From this we want to deduce that $$a\cdot b=0.$$
Square both sides of $$|a+b|=|a-b|$$ to get $$(a+b)\cdot(a+b)=(a-b)\cdot(a-b).$$ Expanding these gives $$a\cdot a+2a\cdot b+b\cdot b=a\cdot a-2a\cdot b+b\cdot b\implies a\cdot b=-a\cdot b,$$ or that $$a\cdot b=0,$$ as desired.
Finally, you may explicitly change this into coordinates by setting $$a=(a_1,a_2),\,b=(b_1,b_2).$$