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 ?

$angle(A,x,B)=angle(C,x,D)$

Is there a known formula that gives these points ?

deiluefniwf
2022-09-25
Answered

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 ?

$angle(A,x,B)=angle(C,x,D)$

Is there a known formula that gives these points ?

You can still ask an expert for help

Klecanlh

Answered 2022-09-26
Author has **11** answers

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:

$x=\frac{\pm \sqrt{(-A+Bt+Ct-D{)}^{2}-4(1-t)(AD-BCt)}-A+Bt+Ct-D}{2(t-1)}.$

Anyway, this gives you the points you're looking for (parametrized by $t\in \mathbb{R}$).

$\frac{x-A}{x-B}/\frac{x-C}{x-D}\in \mathbb{R}.$

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:

$x=\frac{\pm \sqrt{(-A+Bt+Ct-D{)}^{2}-4(1-t)(AD-BCt)}-A+Bt+Ct-D}{2(t-1)}.$

Anyway, this gives you the points you're looking for (parametrized by $t\in \mathbb{R}$).

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