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.
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:
$\displaystyle \overrightarrow{AC}=\overrightarrow{OC} - \overrightarrow{OA}$
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.
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 $\displaystyle O$?
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 $\displaystyle \overrightarrow{AB} = (-2,5)-(3,4)=(-5,1)$
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.
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