# You 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 ?

You have a quadrilateral ABCD. I want to find all the points x inside ABCD such that
$angle\left(A,x,B\right)=angle\left(C,x,D\right)$
Is there a known formula that gives these points ?
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Klecanlh
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
$\left(x-A\right)\left(x-D\right)=t\left(x-B\right)\left(x-C\right).$
For any given $t$ it is a quadratic equation for $x$, so we can solve it; the solution doesn't look very pretty:
$x=\frac{±\sqrt{\left(-A+Bt+Ct-D{\right)}^{2}-4\left(1-t\right)\left(AD-BCt\right)}-A+Bt+Ct-D}{2\left(t-1\right)}.$
Anyway, this gives you the points you're looking for (parametrized by $t\in \mathbb{R}$).