Show that diagonals in a diamond (rhombus) are orthogonal

[Petrus]

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,
\(\displaystyle |\pi\rangle\)

 

[MarkFL]

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

Are you talking about how:

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

If so, this is because:

\(\displaystyle \overline{AD}=\overline{BC}\)

 

[Petrus]

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

Are you talking about how:

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

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]

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 [tex]\overrightarrow {AB} [/tex] as the vector from A to B and [tex]\|\overrightarrow {AB}\| [/tex] as the length of the vector.

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

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

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