Take two fixed points $X$ and $Y$ on $AC$, and construct the corresponding $X_1, X_2, Y_1, Y_2$. Let the circumcircle of $BX_1X_2$ and the circumcircle of $BY_1Y_2$ meet again at point $T$. Then it suffices to prove that for any point $Z$ on ray $AC$, the corresponding $Z_1, Z_2$ are concyclic with $B, T$; in this case the circumcenter of $BZ_1Z_2$ will lie on the perpendicular bisector of segment $BT$.

(Note: If the points $B$ and $T$ coincide, then the circumcircle of $\triangle BX_1X_2$ and the circumcircle of $\triangle BY_1Y_2$ are tangent at point $B$. In this case the problem becomes proving that the circumcircle of $\triangle BZ_1Z_2$ is also tangent to the circumcircle of $\triangle BX_1X_2$ at point $B$.)

Note that, since $XX_2 \parallel YY_2 \parallel ZZ_2$, we have
$$X_2Y_2 : Y_2Z_2 = XY : YZ = X_1Y_1 : Y_1Z_1.$$
Also $\angle Y_1TY_2 = \angle Y_1BY_2 = \angle X_1BX_2 = \angle X_1TX_2$,
so $\angle Y_1TX_1 = \angle X_2TY_2$. On the other hand, $\angle TY_1X_1 = \angle TY_2X_2$, so
$$\triangle TX_1Y_1 \sim \triangle TX_2Y_2.$$
Combining the above, the four points $TX_1Y_1Z_1$ and the four points $TX_2Y_2Z_2$ are similar. Therefore
$$\angle Z_1TX_1 = \angle Z_2TX_2,$$
hence
$$\angle Z_1TZ_2 = \angle X_1TX_2 = \angle X_1BX_2.$$
That is, $BTZ_1Z_2$ are concyclic, as desired.