# Show that diagonals in a diamond (rhombus) are orthogonal

#### Petrus

##### Well-known member
Hello MHB,
I am working with a exemple that I think they got some incorrect.
Exemple 10.Show that diagonals in a diamond(romb) is orthogonal.

I understand that $$\displaystyle AC•BD=0$$ cause of dot product and if it's orthogonal the angle is $$\displaystyle \frac{\pi}{2}$$
I understand all the part until the step before the last one Edit: last part got cut in picture, it should be: $$\displaystyle |BC|^2-|AB|^2=0$$

Regards,
$$\displaystyle |\pi\rangle$$

Last edited:

#### MarkFL

Staff member
Re: Show that diagonala in a diamond(romb) is orthogonal

$$\displaystyle \overline{AB}\cdot\left(\overline{AD}-\overline{BC} \right)=0$$ ?

If so, this is because:

$$\displaystyle \overline{AD}=\overline{BC}$$

#### Petrus

##### Well-known member
Re: Show that diagonala in a diamond(romb) is orthogonal

$$\displaystyle \overline{AB}\cdot\left(\overline{AD}-\overline{BC} \right)=0$$ ?

If so, this is because:

$$\displaystyle \overline{AD}=\overline{BC}$$
Hello Mark!
Thanks for the fast respond! Now evrything make sense! I did confuse myself and thought they did rewrite $$\displaystyle AB•AD=AB•(AD-BC)$$ but they did rewrite $$\displaystyle AB•AD-BC•AB=AB(AD-BC)$$ and indeed $$\displaystyle AD=BC$$ so we could also rewrite that as $$\displaystyle AB(BC-BC)$$ or $$\displaystyle AB(AD-AD)$$. I did not see that.. Thanks alot evrything make sense now!

Regards,
$$\displaystyle |\pi\rangle$$

#### Plato

##### Well-known member
MHB Math Helper
Hello MHB,
I am working with a exemple that I think they got some incorrect.
Exemple 10.Show that diagonals in a diamond(romb) is orthogonal.
There is little new in my post except to point out some advantages of vector notation.

Use $$\overrightarrow {AB}$$ as the vector from A to B and $$\|\overrightarrow {AB}\|$$ as the length of the vector.

Now if $$\overrightarrow {AB} ~\&~\overrightarrow {AD}$$ are adjacent sides of a convex quadrilateral, then the diagonals are $$\overrightarrow {AB}+\overrightarrow {AD}~\&~\overrightarrow {AB}-\overrightarrow {AD}$$.

Note that $$\left( {\overrightarrow {AB} + \overrightarrow {AD} } \right) \cdot \left( {\overrightarrow {AB} - \overrightarrow {AD} } \right) = {\left\| {\overrightarrow {AB} } \right\|^2} - {\left\| {\overrightarrow {AD} } \right\|^2}$$.

But in a rhombus those two sides are equal length, giving us zero.