TITLE: On Quadrilaterals
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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 $t\in\mathbb{R}$).