# Thread: vector Proof - diagonals of a kite

1. ## 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.

2. 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.

3. 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 $O$?

4. 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 $\overrightarrow{AB} = (-2,5)-(3,4)=(-5,1)$

5. 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.

6. 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.

7. Here is another definition of a kite

8. 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.

9. ## 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 $AC$.