Notice that the quadrilateral also must be cyclic. In this case, denote by $ M$, $ N$ the midpoints of the diagonals $ BD$, $ AC$, respectively, and by $ P$ the intersection point of the diagonals of $ ABCD$. Since the lines $ BD$, and $ BM$ are isogonal with respect to angle $ ABC$, by Steiner's theorem, $ \frac{AB^{2}}{BC^{2}}=\frac{AP}{PC}.$ Similarly, the lines $ DB$ and $ DM$ are isogonal with respect to angle $ CDA$, and therefore $ \frac{AD^{2}}{DC^{2}}=\frac{AP}{PC}.$ Hence $ AB/BC=AD/DC$; in other words, the quadrilateral $ ABCD$ is harmonic. Now, since the pencil $ B(B, A, D, C)$ is harmonic, by intersecting it with the line $ AC$, we deduce that $ T$ (the intersection point of the tangent at $ B$ to the circumcircle of $ ABCD$ and line $ AC$) is the harmonic conjugate of $ P$ with respect to segment $ (AC)$. On other hand, the pencil $ D(D, A, B, C)$ is harmonic, and thus the tangent at $ D$ to the circumcircle of $ ABCD$ passes through $ T$. It is now well-known that if $ XYZ$ is the tangential triangle of a given triangle $ ABC$, the lines $ AX$, $ BY$, $ CZ$ are the symmedians of $ ABC$. According to this fact, and to the collinearity of the points $ A$, $ C$, and $ T$, the diagonal $ AC$ is the $ A$-symmedian of triangle $ ABD$, and simultanously the $ C$-symmedian of triangle $ BCD$. Therefore, the reflections of $ AC$ in the bisectors of angles $ A$, and $ C$ of the cyclic quadrilateral $ ABC$ pass through the midpoint of $ BD$. This completes our proof.

I have another solution, which uses the following known fact: considering the projective plane with the circumcircle of $ ABC$ as reference, the symmedians of $ ABC$ are obtained connecting the corresponding vertex to the dual of the opposite side. Then the problem asks to prove that if $ AD$ is the $ A$-symmedian of $ ABC$ then $ BC$ is the $ B$ symmedian of $ ABD$. In order to prove that, let $ P$ be the dual point of $ BC$ and $ Q$ be the dual point of $ AD$. Then $ P$ belongs to $ AD$ and the result follows from the duality theorem: the dual of $ P$, $ BC$ passes through the dual of $ AD$, $ Q$.