Given an angle $\angle ABC$, where $B$ is the vertex, and segments $BA$ and $BC$ are not congruent, how do you bisect the angle compass only?

Question

Im looking for a valid Mohr-Mascheroni (compass-only) construction in Euclidean geometry. What Im looking for isnt covered in any of my books and I cant find it online. Im trying to bisect an arbitrary angle defined by three points in space, unconnected by drawn line segments, but unfortunately having non-congruent legs, thus a simple arc bisection will not suffice. I can brute force a construction if Im willing to do the work but Im wondering if there is a simpler method, say, under ten circles.

Answer 1

Here is a possible construction. We start with the blue points $A,B,C$, and want to construct using only the compass a point on the angle bisector of $\widehat{ABC}$:

Using https://en.wikipedia.org/wiki/Mohr%E2%80%93Mascheroni_theorem#Intersection_of_a_line_and_a_circle_(construction_#4) (the second case) we find the point $D$ of the intersection of the line $AB$ with the circle $\mathcal C(B, BC)$, centered in $B$ with radius $BC$. (Here, we construct the point $D$ such that $B$ is on the segment $AD$.)

Then we only need a point on the parallel through $B$ of the line $CD$, which is the needed angle bisector. Construct now the point $F$ such that $BFCD$ is a parallelogram. (Intersect the circles $\mathcal C(B,CD)$ and $\mathcal C(C,BD)$ and consider the "right intersection point".)