I've been looking at spiral similarities as a problem solving tactic, which I have found surprisingly versatile (within synthetic geometry), however I have had trouble with a specific configuration.
I should mention, I've been using the following lemma to find this point:
Define $P$ to be the intersection of $AC$ and $BD$ (i.e. $P = AC\cap BD$). Circles $(ABP)$ and $(CDP)$ meet at $O$ ($O\neq P$); thus, $O$ is the center of the spiral similarity which takes $AB$ to $CD$.
But what do you do when one of the points ($A$, $B$, $C$ or $D$) is $P$? I know there should be a spiral similarity for them (because I heard there exists a unique spiral similarity for any two segments that are not parallel), but it is difficult because it means one of the circles above is only defined using two points, meaning I cannot fix the circle...
What do I need this for? There are problems where I think two triangles are similar, and I can't think of any other way of showing this than through spiral similarity. To do this I need to prove that some point is the spiral centre.
Thanks for any help you can provide.