I recently solved this exercise from Hartshorne's Classical Geometry.
Given three lines $a$, $b$, $c$ through a point $O$, show that there exists a unique fourth line $d$ such that $$\sigma_c\sigma_b\sigma_a=\sigma_d,$$ where $\sigma$ denotes the reflection in a given line.
In the diagram below, I take $A\in a$, let $B$ be the reflection of $A$ across $b$, $C$ be the reflection of $B$ across $c$, and then let $d$ be the perpendicular bisector of $AC$. I was able to prove $\sigma_c\sigma_b\sigma_a=\sigma_d$, which I will be happy to include if someone asks. Anyway, based on that, I know that $\sigma_d$ fixes $O$, and thus passes through $O$.
My follow up question to this exercise is, is the angle between $a$ and $d$ congruent to the angle between $b$ and $c$? I'm curious to know, because if so, I believe this implies that given three reflection axes through a point, we can rotate two of them so that one coincides with the third, hence cancelling, and the image of the first rotated line will be the desired $d$.
I have a hunch it is true based on an earlier question, but this seems like a different way of proving it. I've included a picture with which angles I know are congruent, and I assume I'm working in a Euclidean plane. Thanks for any insight.