# Thread: "A parallelogram is a rhombus if and only if its diagonals are at right angles"

1. ## "A parallelogram is a rhombus if and only if its diagonals are at right angles"

I proved that ||a+b|| = ||a-b|| if and only if a.b = 0

However the next part says using this explain why a paralleogram is a rhombus if and only if its diagonals are at right angles..

I can't figure this out?

2. This is a neat problem.
If $\vec{a}~\&~\vec{b}$ are consecutive sides of a parallelogram then $\vec{a}+\vec{b}~\&~\vec{a}-\vec{b}$ are its diagonals.
Now $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b})=\vec{a}\cdot \vec{a}-\vec{b}\cdot\vec{b}$.

So if the parallelogram is a rhombus then $\left\| \vec{a} \right\|=\left\| \vec{b} \right\|$.
What does that imply?

What is the converse?

3. Originally Posted by Plato
This is a neat problem.
If $\vec{a}~\&~\vec{b}$ are consecutive sides of a parallelogram then $\vec{a}+\vec{b}~\&~\vec{a}-\vec{b}$ are its diagonals.
Now $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b})=\vec{a}\cdot \vec{a}-\vec{b}\cdot\vec{b}$.

So if the parallelogram is a rhombus then $\left\| \vec{a} \right\|=\left\| \vec{b} \right\|$.
What does that imply?

What is the converse?
I understand exactly what you've done so far, but still I don't know what it implies!!!

If the paralleogram is a rhombus then it's diagonals must be at right angles. This means a.b = 0

Now what?!

4. I frankly do not know what to say to that.
$\left\| {\vec{a}} \right\|=\left\| {\vec{b}} \right\|$ if and only if $\vec{a} \cdot\vec{a}=\vec{b} \cdot\vec{b}$