vector Proof - diagonals of a kite

Printable View

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

1 Attachment(s)

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.
June 1st 2009, 03:22 AM
Plato

Here is another definition of a kite
June 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.
June 1st 2009, 11:25 AM
Grandad

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 $AC$ .

Grandad
June 1st 2009, 02:04 PM
Enedrox

I assumed that any kite were symmetric (and so did my teacher) but obviously there aren't. I'll have to tell him that.

And thanks Grandad for showing that the diagonals are perpendicular in a symmetric kite.