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I have a quadrilateral ABCD. I want to find all the points x inside ABCD such that $angle(A,x,B)=angle(C,x,D)$

Is there a known formula that gives these points ?

Example:

ABCD is a rectangle. Let $x_1=mid[A,D]$ and $x_2=mid[B,C]$. The points x are those lying on the line that passes through $x_1$ and $x_2$.

But I want a formula for arbitrary quadrilaterals.

Thank you.

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    My first thought is to take the known result for a rectangle and try applying a transformation that preserves lines and angles to map the rectangle onto another quadrilateral. I suspect, though, that such transformations are limited (I think they only have 4 degrees of freedom, so would be determined by the images of 2 given points) and won't get from a rectangle to any general quadrilateral.2011-05-10

2 Answers 2

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If you understand $A,B,C,D,x$ as complex numbers then your condition is $\frac{x-A}{x-B}/\frac{x-C}{x-D}\in\mathbb{R}.$ Let us denote that real number $t$, i.e. you have equation $(x-A)(x-D)=t(x-B)(x-C).$ For any given $t$ it is a quadratic equation for $x$, so we can solve it; the solution doesn't look very pretty: $x = \frac{\pm\sqrt{(-A+B t+C t-D)^2-4 (1-t) (A D-B C t)}-A+B t+C t-D}{2 (t-1)}.$ Anyway, this gives you the points you're looking for (parametrized by $t\in\mathbb{R}$).

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For a "formula" we would first have to discuss what constitutes an answer, but I made a picture to make it clear that the condition "inside" is not a very natural one.

enter image description here

I used geogebra. Note that when the curve crosses the line CD or AB , you do not have equal angles anymore, instead the smaller angles sum to 180 degrees, but it is fine again when the curve crosses again.

Furthermore, note that if you move the vertex A slightly, the part of the curve that passes through $A$ and $B$ becomes detached and formes a little oval curve.

If you want to see an equation: $\frac{((a1 - x) (b1 - x) + (a2 - y) (b2 - y))}{\sqrt{((a1 - x)^2 + (a2 - y)^2) ((b1 - x)^2 + (b2 - y)^2)}} = \frac{((c1 - x) (d1 - x) + (c2 - y) (d2 - y))}{\sqrt{((c1 - x)^2 + (c2 - y)^2) ((d1 - x)^2 + (d2 - y)^2)}}$

It does not get better if you square it.

Non-convex quadrilaterals do not look different, they also can pass from an S-curve to a little oval plus another branch.

enter image description here

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    @user3749 I want to point out that you have voted neither for my answer nor for the other answer yet. You *said* you would be looking up the connection of cosinus and inner product and this is exactly what I used. Angles equal implies cosinus equal. And in any case, I have to go to bed.2011-05-10