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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Do you know the properties of inscribed angles? –  Phira May 10 '11 at 14:58
Or the expression for the cosinus using the inner product? –  Phira May 10 '11 at 15:06
@user9325: Sorry I don't understand your question. You mean if I have a constraint about the angles (A,x,B) and (C,x,D) ? Nothing appart from them being equal. –  user3749 May 10 '11 at 15:08
I'm having a hard time coming up with something that would work even for "arrowheads"... –  Guess who it is. May 10 '11 at 15:14
@user3749: I want to know what you know about geometry. Do you know the concepts that I have mentioned? –  Phira May 10 '11 at 15:46

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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+1 for the parametric formula. –  Phira May 10 '11 at 20:58
Thank you very much. –  user3749 May 11 '11 at 0:35

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.

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.

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What is a formula? math.stackexchange.com/questions/38155/… :-) –  lhf May 10 '11 at 16:55
@lhf: Well, yes, so the point is that it is not too hard to find a polynomial equation for the curve (using coordinates and equality of cosines expressed with inner products), but the degree would be quite high and the question of being inside even more intractable than the question of an explicit formula. –  Phira May 10 '11 at 17:01
My visualization skills are failing me now, but it looks as if the "arrowhead" case might be an even more perverse configuration... –  Guess who it is. May 10 '11 at 18:08
@I.J. Kennedy: Geogebra has a function that draws the locus of points when a certain point moves along a line. In my first picture you can see the point $E$ that moves along the bisector of $AB$. $E$ was the center of a circle through $A$ and then I drew a circle with corresponding angles through $CD$, then I defined the intersection of the two circles as $I,J$ and let geogebra draw the locus of $I,J$ when $E$ moves along the bisector. –  Phira May 10 '11 at 20:43
@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. –  Phira May 10 '11 at 23:45