# Easier way to calculate this point besides line intersection?

Given are all points except E, plus |AF| = |DC|. Considering that the lines AB and FE, as well as BC and ED are parallel, is there an easier way to calculate E? Maybe some relation with B?

I'd like to express the coordinates from E as short as possible and therefore I didn't want to calculate the intersection between the lines FE and ED.

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Is $|AB| = |CD|$? – Mikael Öhman Aug 10 '11 at 17:58
@Mikael, not necessarily, so no – Clash Aug 10 '11 at 18:02
I'm afraid I don't think there is. I don't understand your concern about writing the coordinates as short as possible; work out the intersection point the long way, and then simplify whatever you get as much as you can, and that's as short as it will go. – Billy Aug 10 '11 at 18:34
btw, I forgot to mention |AF| = |DC|, but I think this wouldn't help either. @Billy these formulas are inserted into a file, and I wanted them to be as short and as easy to understand as possible. Thanks for the help guys! – Clash Aug 10 '11 at 18:52
I had an answer but I must have missed the "besides line intersection" part in the title. No - I don't think there is an easier way than line intersection, because that method is very easy and any other way seems like it'd be a roundabout version of just that. – anon Aug 10 '11 at 19:27

Here is an answer without explanation.

$\vec{E} = \vec{B}+\frac{1}{1+\frac{\vec{CD}\cdot\vec{AF}}{\vec{AF}\cdot\vec{AF}}}(\vec{AF}+\vec{CD})$

Here is the explanation:

First, translate so that $B$ is at the origin. This will simplify notation. At the end, we will translate back, obtaining the "$\vec{B}+$" in the formula.

Sketch the vectors $\vec{AF}$ and $\vec{CD}$ with their heads at $E$. Sketch direct lines from their tails to $B$. The right angles and the fact that $|AF|=|CD|$ imply that this quadrilateral is a kite. Moreover, the diagonal out of $B$ is spanned by the vector $(\vec{AF}+\vec{CD})$. So $\vec{E}$ is some scalar multiple of this: $\vec{E}=t(\vec{AF}+\vec{CD})$.

The conditions that have been laid out demand that $$proj_{\vec{AF}}\vec{E}=\vec{AF}$$ That means $\frac{\vec{AF}\cdot\left(t(\vec{AF}+\vec{CD})\right)}{\vec{AF}\cdot\vec{AF}}\vec{AF}=\vec{AF}$, or rather $\frac{\vec{AF}\cdot\left(t(\vec{AF}+\vec{CD})\right)}{\vec{AF}\cdot\vec{AF}}=1$. This implies $$t=\frac{\vec{AF}\cdot\vec{AF}}{\vec{AF}\cdot\vec{AF}+\vec{AF}\cdot\vec{CD}}$$ which simplifies to the coefficient in the solution.

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