Find points A and B from their angle bisectors I need to find points $A$ and $B$ in $ℝ^2$, when I only know three points $X_1, X_2, X_3$ and angle bisectors of the angles $A X_1 B$, $A X_2 B$ and $A X_3 B$. How do I do this geometrically?

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If I used only two points – $X_1$ and $X_2$ – and their respective lines, I'm able to find a point $B'$ for every $A'$, so that it fulfils the condition. This means it has infinite solutions.

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For the two points the solution was pretty easy – I just needed draw the line $AX_1$ and then another one that is axially symmetric to it, then repeat for $X_2$. $B$ is at the intersection of the two axially symmetric lines.
However this approach isn't of much use when I use all the three points, as the lines usually don't meet at a single point.

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There's always one point A' such that the three lines do meet in one point. This is the solution, however I wasn't able to find it otherwise than moving it here and there using trial and error.
 A: In projective geometry and using homogeneous coordinates, you can express an angular bisector like this:
$$(A,Y_1;I,J)_{X_1} = (Y_1,B;I,J)_{X_1}$$
The parentheses denote cross ratios. $X_1$ is the point from which ous the cross ratio is seen; you might also consider this as the cross ratio of the lines from the given points to $X_1$. $Y_1$ is any other point on the given bisector; it might be a point at indfinity describing the direction. $I$ and $J$ are the ideal circle points with homogeneous coordinates $I=(1:i:0)$ and $J=(1:-i:0)$. The whole cross ratio is proportional to the exponent of the angle, via Laguerre's formula. Expanding the above and using square brackets for determinants, you get
$$[A,I,X_1][Y_1,J,X_1][Y_1,J,X_1][B,I,X_1]=[A,J,X_1][Y_1,I,X_1][Y_1,I,X_1][B,J,X_1]$$
From this you can conclude that $B$ lies on the line
$$[A,I,X_1][Y_1,J,X_1][Y_1,J,X_1](I\times X_1)-[A,J,X_1][Y_1,I,X_1][Y_1,I,X_1](J\times X_1)$$
which is just a complicated way to express what you get by reflecting the line $AX_1$ in the line $X_1Y_1$. If you want to, you can get rid of complex coordinates at this point, since they can be collected into a single common factor which may be discarded for homogeneous coordinates.
You can do the same computation (or merely adapt the result) for the other two pairs $(X_2,Y_2)$ and $(X_3,Y_3)$ and obtain two more lines. These three lines will be concurrent (and $B$ their point of intersection) iff their determinant is zero. So at this point you have a single equation in $A$ which is zero iff $A$ leads to a valid $B$. The point $A$ occurs of degree $3$ in this equation, so $A$ has to lie on a given cubic curve, its position determined by the $X_i$ and $Y_i$.
