Let $XYZ$ be a satisfactory triangle. Then the vertices $Y$ and $Z$ must lie in the same half-plane with the boundary line $AB$. Denote by $Y'$ the reflection of $Y$ through the line $AB$. Due to the presumed similarity, the angles $XAZ$ and $BYX$ are congruent (Fig. 1) and hence $|\angle BAZ| = |\angle BY'Z|$ as well. Using the well known inscribed angles property we conclude that the circumcircle of $\triangle ABZ$ passes not only through the point $Y$, but also through the point $Y'$. The line $AB$ (as a perpendicular bisector of the chord $YY'$) passes through the centre $O$ of the circle $k$ and thus the chord $AB$ is a diameter of $k$. Since the segment $AB$ is fixed, the circle $k = ABYZ$ is common for all satisfactory triangles $XYZ$ and the midpoint $M$ of $YZ$ must lie in the interior of $k$. Since the both angles $OMZ$ and $OMY$ are right (Fig. 2), the (lesser) angles $AMO$ and $BMO$ are acute and thus the point $M$ must lie in the intersection of the exteriors of Thales' circles with diameters $AO$ and $BO$. In what follows we will show the both derived necessary conditions determine the locus of all the possible midpoints $M$.

Fig. 1

Fig. 2

So, let $M$ be any point in the interior of $k$ for which the both angles $AMO$ and $BMO$ are acute (i.e. $M$ lies in the exteriors of the circles with diameters $AO$ and $BO$). Consider a chord of $k$ which passes through $M$ perpendicularly to $OM$. This chord does not intersect the diameter $AB$, because of the acute angles $AMO$ and $BMO$. Thus the endpoints of the chord with the midpoint $M$ can be denoted as $Y$ and $Z$ so that $A$, $B$, $Y$, $Z$ lie on $k$ in this order. If $Y$ reflects to $Y'$ through the diameter $AB$ and if $X$ denotes the intersection point of the segments $AB$ and $Y'Z$, then the triangles $XBY$ and $XZA$ are similar as required (by theorem AA). This completes the solution.

Fig. 1

Fig. 2

Conclusion. The locus under consideration is the interior of the highlighted region bounded by the three circles with diameters $AB$, $AO$ and $BO$, where $O$ denotes the midpoint of segment $AB$ (Fig. 3).

Fig. 3