# vector Proof - diagonals of a kite

• May 30th 2009, 09:53 PM
Enedrox
vector Proof - diagonals of a kite
Another vector proving problem:

Take any kite shape ABCD. How do I prove that the diagonals (AC and BD) are perpendicular for any kite?

I beleive it should include the dot-product of the diagonals except I dunno how to prove this for any kite.
• May 30th 2009, 11:47 PM
earboth
Quote:

Originally Posted by Enedrox
Another vector proving problem:

Take any kite shape ABCD. How do I prove that the diagonals (AC and BD) are perpendicular for any kite?

I beleive it should include the dot-product of the diagonals except I dunno how to prove this for any kite.

1. Express the diagonals as differences of stationary vectors:

$\overrightarrow{AC}=\overrightarrow{OC} - \overrightarrow{OA}$
and
$\overrightarrow{BD}=\overrightarrow{OD} - \overrightarrow{OB}$

Then prove that

$\overrightarrow{AC} \cdot \overrightarrow{BD} = 0$

2. Symmetric kite:

Additional to the proof of the orthogonality you must show that one diagonal is the bisector of the other one.
• May 31st 2009, 01:41 AM
Enedrox
sorry, but I still don't know how to prove that
$\overrightarrow{AC} \cdot \overrightarrow{BD} = 0$

I get:
$\overrightarrow{AC} \cdot \overrightarrow{BD} = (\overrightarrow{OC} - \overrightarrow{OA}) \cdot (\overrightarrow{OD} - \overrightarrow{OB})$
$\overrightarrow{AC} \cdot \overrightarrow{BD} = \overrightarrow{OC} \cdot \overrightarrow{OD} - \overrightarrow{OC} \cdot \overrightarrow{OB} - \overrightarrow{OA} \cdot \overrightarrow{OD} - \overrightarrow{OA} \cdot \overrightarrow{OB}$

however what does that tell me?

Is there a way of proving that the diagonals $\overrightarrow{AC}$ and $\overrightarrow{BD}$ are perpendicular without involving another point $O$?
• May 31st 2009, 06:52 AM
earboth
Quote:

Originally Posted by Enedrox
sorry, but I still don't know how to prove that
$\overrightarrow{AC} \cdot \overrightarrow{BD} = 0$

I get:
$\overrightarrow{AC} \cdot \overrightarrow{BD} = (\overrightarrow{OC} - \overrightarrow{OA}) \cdot (\overrightarrow{OD} - \overrightarrow{OB})$
$\overrightarrow{AC} \cdot \overrightarrow{BD} = \overrightarrow{OC} \cdot \overrightarrow{OD} - \overrightarrow{OC} \cdot \overrightarrow{OB} - \overrightarrow{OA} \cdot \overrightarrow{OD} - \overrightarrow{OA} \cdot \overrightarrow{OB}$

however what does that tell me?

Is there a way of proving that the diagonals $\overrightarrow{AC}$ and $\overrightarrow{BD}$ are perpendicular without involving another point $O$?

Sorry, I should have mentioned that O is refering to the origin of the coordinate system.

For instance: If you have the point A(3,4) then the vector $\overrightarrow{OA} = (3, 4)$; and if you have a point B(-2,5) then the vector $\overrightarrow{OB} = (-2, 5)$

Consequently the vector $\overrightarrow{AB} = (-2,5)-(3,4)=(-5,1)$
• May 31st 2009, 01:44 PM
Enedrox
That's all good, except that by giving an example, I'm no longer proving this for any kite..

I somehow have to use what all kite have in common to prove that the diagonals are perpendicular.
• May 31st 2009, 09:42 PM
earboth
Quote:

Originally Posted by Enedrox
That's all good, except that by giving an example, I'm no longer proving this for any kite..

I somehow have to use what all kite have in common to prove that the diagonals are perpendicular.

In my opinion you can't prove a definiton:

As far as I'm informed a kite is a quadrilateral whose diagonals are perpendicular. And that's what all kites have in common.

With this definition and the calculations I gave in my previous posts you can prove that a square, a rhombus, some very special trapezoids are kites too, but you can't prove that a kite is a kite.

If you want to prove that a quadrilateral is a kite you must know some properties of the (unknown) quadrilateral. Then use the calculations to prove if the diagonals are perpendicular or not.
• Jun 1st 2009, 03:22 AM
Plato
Here is another definition of a kite
• Jun 1st 2009, 06:46 AM
earboth
Quote:

Originally Posted by Plato
Here is another definition of a kite

Thanks for the link. But I know this kind of quadrilateral as symmetric kite, which is a special case of the definition I used in my previous posts.
• Jun 1st 2009, 11:25 AM
Diagonals of a kite
Hello everyone -

I have always used Plato's definition of a kite. So the vector proof you want goes like this.

Suppose that in the quadrilateral ABCD, $|AB|= |BC|$ and $|AD| = |DC|$, and let $M$ be the mid-point of the diagonal $AC$. Then:

$\vec{AC} = \vec{AB} + \vec{BC}$

and $\vec{MB} = \vec{MA} + \vec{AB}$

$= -\tfrac12\vec{AC}+ \vec{AB}$

$= \tfrac12(\vec{AB} - \vec{BC})$

$\Rightarrow \vec{MB}.\vec{AC} = \tfrac12(\vec{AB}-\vec{BC}).(\vec{AB}+\vec{BC}) = \tfrac12(|AB|^2 - |BC|^2) = 0$

$\Rightarrow \vec{AC}$ and $\vec{MB}$ are perpendicular.

Similarly in $\triangle ADC, \vec{MD}$ and $\vec{AC}$ are perpendicular.

$\Rightarrow BMD$ is a straight line - namely, the diagonal $BD$, which is therefore perpendicular to the diagonal $AC$.