You have a quadrilateral ABCD. I want to find all the points x inside ABCD such that a n g l e ( A , x , B ) = a n g l e ( C , x , D )Is there a known formula that gives these points ?

Question

You have a quadrilateral ABCD. I want to find all the points x inside ABCD such that  a  n  g  l  e  (  A  ,  x  ,  B  )  =  a  n  g  l  e  (  C  ,  x  ,  D  )Is there a known formula that gives these points ?

Klecanlh · Accepted Answer

If you understand A,B,C,D,x as complex numbers then your condition is            x      −      A              x      −      B            /              x      −      C              x      −      D        ∈      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&#039;t look very pretty:  x  =            ±              (        −        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&#039;re looking for (parametrized by   t  ∈      R  ).

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 ?

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