Suppose we have three concurrent lines $g,h,k$ in the Euclidean plane which meet at a point $P\in g\cap h\cap k.$ Moreover, let $K$ be some circle with center $P$ and some radius $r>0$.
I would like to construct, with ruler and compass, all triangles with circumcircle $K$ and such that $g,h,k$ become the perpendicular bisectors of the sides of the triangle.
My ideas so far: I proved that the composition of reflections $s_g\circ s_h\circ s_k$ is again a reflection in some line through $P$ (where $s_g$ denotes the reflection in the line $g$ etc.). If we permute the order of the three reflections ($s_g,s_h, s_k$), we get again a reflection but in a different line through $P$.
Using this observation, I came up with the following idea for a construction. Suppose, we reflect some point $A\in K\cap g$ successively at $k$ and then at $h$. We obtain some point $A'$ (This would correspond to the point $A'=s_h\circ s_k\circ s_g(A)$, i.e. A' is the reflection of $A$ at some line $d$ through $P$). When we construct the bisector of $A A'$, we get the line of reflection $d$ of $s_h\circ s_k\circ s_g$. Let $Q\in d\cap K$. If we reflect the point $Q$ successively at $g,k,h$, we obtain a triangle, which meets all criteria.
If we do the same with all other points, we get in total two triangles which meet all criteria. But I do not know how to prove that there are no further such triangles. Moreover, it seems that my construction is somehow lengthy. Is it possible to construct the triangles more elegant?
I am very grateful for your help! Best wishes!
Your solution is great. If $g,h,k$ are bisectors of $BC,CA,AB$ respectively, since $S_h\circ S_g\circ S_k(A)=A$ and $S_h\circ S_g\circ S_k(P)=P$, and we know that $S_h\circ S_g\circ S_k$ is a line reflection $S_d$, we conclude that it must be $d=PA$. Thus $A$ must be an intersection of $d$ and the circumcircle, so there are only two possibilities for $A$, and hence only two solutions.
For another solution you can use the following facts: Let $ABC$ be a triangle, then:
Fact 1. Bisector of $\angle A$ and bisector of $BC$ intersect on the circumcircle.
Denote by $A'$ the point from Fact 1, and similarly define $B'$ and $C'$.
Fact 2. $AA'\perp B'C'$.
If $A''$ is another intersection of bisector of $BC$ with the circumcircle, and $B''$ and $C''$ are similarly defined.
Fact 3. On the circumcircle, the points are arranged as follows: $A',B'',C',A'',B',C''$.
Construction. Denote the intersections of $g$ with the cicumcircle by $A',A''$, of $h$ by $B',B''$ and of $k$ by $C',C''$, arranged as in Fact 3. By Fact 2, the heights in triangle $A'B'C'$ intersect the circumcircle in desired $A,B,C$. The second solution you get if you consider $A''B''C''$ instead of $A'B'C'$.