Find point such that the triangle it makes with two distinct line segments are similar

I was working on a proof about dilative rotations and, while I managed to prove a solution, I couldn't prove its uniqueness. Then I had the idea of proving that there was only one point such that those two mentioned triangles are similar. So given two pairs of points, how do I find a point C such that the this is the case? I tried using a bit of vector algebra and came up with the set of equations:

$\frac{\left\lVert C-B \right\rVert}{sin(\alpha)} = \frac{\left\lVert C-A \right\rVert}{sin(\beta)} = \frac{\left\lVert B-A \right\rVert}{sin(\theta)}$

$\frac{\left\lVert C-B' \right\rVert}{sin(\alpha)} = \frac{\left\lVert C-A' \right\rVert}{sin(\beta)} = \frac{\left\lVert B'-A' \right\rVert}{sin(\theta)}$

, where B-A is a one line segment(or vector), B'-A' is the other one and $\frac{\left\lVert C-B' \right\rVert}{\left\lVert C-B \right\rVert} = \frac{\left\lVert C-A' \right\rVert}{\left\lVert C-A \right\rVert} = \frac{\left\lVert A'-B' \right\rVert}{\left\lVert A-B \right\rVert} = r$

Problem is, this seems to get algebraically complicated really fast. Do you have any smarter ideas on how to go about this?

The point you are looking for of called the center of the spiral similarity. To prove uniqueness, observe that if $X'$ is another center, then $ABPX'$ and $DCPX'$ are also cyclic (I'm following the Wikipedia notation), since two distinct circles have at most two intersection points, and $X\neq P\neq X'$, we get that $X=X'$, which is what we wanted.

The proof may have several cases depending whether the lines $AC$ and $BD$ intersect inside the segments of not, but are all similar (you can avoid card using directed angles if you know then, but that's not necessary).