MHB 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 $$AC•BD=0$$ cause of dot product and if it's orthogonal the angle is $$\frac{\pi}{2}$$
I understand all the part until the step before the last one

Edit: last part got cut in picture, it should be: $$|BC|^2-|AB|^2=0$$

Regards,
$$|\pi\rangle$$

 

[MarkFL]

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

Are you talking about how:

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

If so, this is because:

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

 

[Petrus]

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

Are you talking about how:

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

If so, this is because:

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

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

Regards,
$$|\pi\rangle$$

 

[Plato2]

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.