Another vector proving problem:

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

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

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- May 30th 2009, 09:53 PMEnedroxvector 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 PMearboth
1. Express the diagonals as differences of stationary vectors:

$\displaystyle \overrightarrow{AC}=\overrightarrow{OC} - \overrightarrow{OA}$

and

$\displaystyle \overrightarrow{BD}=\overrightarrow{OD} - \overrightarrow{OB}$

Then prove that

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

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 AMEnedrox
sorry, but I still don't know how to prove that

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

I get:

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

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

however what does that tell me?

Is there a way of proving that the diagonals $\displaystyle \overrightarrow{AC}$ and $\displaystyle \overrightarrow{BD}$ are perpendicular without involving another point $\displaystyle O$? - May 31st 2009, 06:52 AMearboth
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 $\displaystyle \overrightarrow{OA} = (3, 4)$; and if you have a point B(-2,5) then the vector $\displaystyle \overrightarrow{OB} = (-2, 5)$

Consequently the vector $\displaystyle \overrightarrow{AB} = (-2,5)-(3,4)=(-5,1)$ - May 31st 2009, 01:44 PMEnedrox
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 PMearboth
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 AMPlato
Here is another definition of

*a kite* - Jun 1st 2009, 06:46 AMearboth
- Jun 1st 2009, 11:25 AMGrandadDiagonals 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, $\displaystyle |AB|= |BC|$ and $\displaystyle |AD| = |DC|$, and let $\displaystyle M$ be the mid-point of the diagonal $\displaystyle AC$. Then:

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

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

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

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

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

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

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

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

Grandad - Jun 1st 2009, 02:04 PMEnedrox
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.