We're given a random red line and a point $A$ outside of it.
From the red line, we take a random point $B$ such that the circle $\omega = \odot (A,AB)$ cuts the red line again at $C \neq B$.
Let $D = \odot(B,BA) \cap \odot (A,AB)$ be on the same half plane of the red line as $A$ and let $E$ be the meeting of the perpendicular bisector of $D$ and $C$ with the red line. Show that $AEBD$ is cyclic.
I believe we need a congruence of triangles here. It is clear that $\triangle DAE \cong \triangle CAE$ by SSS. But I feel like I still need a little detail I'm not quite seeing, can you guys help me?
This is trivial by phantom points. Let $E' = \odot (BDA) \cap BC$. You already know $AD=AC$ and $60^\circ = \measuredangle ABD = \measuredangle AED = \measuredangle CEA $. Moreover $\measuredangle E'DA = \measuredangle E'BA = \measuredangle CBA = \measuredangle ACB = \measuredangle ACE'$. Then by AAS congruency, $AE'$ is the perpendicular bisector of $CD$, i.e., $E = E'$, as desired. $\blacksquare$