The requested set of points $C$ is a sum of two circles passing through $B$ without point $B$. Let $M$ be the midpoint of $AC$ and $N$ be the midpoint of $AB$. Let $O$ be the vertex of equilateral triangle $NBO$. Denote the following circles: $o_{1}=(N,NB)$ $o_{2}=(O,OB)$ $o_{3}=(B,BM)$ Let us take such point $C$ that $M$ lies on circle $o_{2}$. Let $P$ be the intersection of $o_{1}$ and $o_{3}$ on the other side of $AB$ than $C$. Obviously, $AP$ is tangent to $o_{3}$, therefore $BP\perp AP$. As $\triangle BON$ is equilateral, also $\triangle BMP$ is equilateral. So the altitude of $\triangle BMP$ from $M$ passes through $N$. Hence, $MN\perp BP$. This implies $AP\parallel BC$. So $BP$ is the length of the altitude of $\triangle ABC$ from $A$. But $BP=BM$, so our point $C$ satisfies the necessary conditions. For a given altitude from $A$ there is only one "good" point $C$ for each side of line $AB$. Therefore, the requested set of points $C$ is a circle $o$ homothetic to $o_{2}$ with respect to $A$ in scale $k=2$ and a circle symmetric to $o$ with respect to $AB$ (excluding point $B$).