Notice that $\angle CFB = 90^{\circ}; BD = DC$ and $\angle BEC = 90^{\circ}; CD = DB$. That gives us $FD = BD = DC$ and $ED = BD = DC$, which is equivalent to $BD = FD = ED = CD$, and we find that $FECB$ is a concyclic quadrilateral (with $D$ being it's center). Let's denote point $H$ as the intersection of $BE$ and $CF$, $\angle HAF = x$ and $\angle HAE = y$. Using that $AEHF$ is concyclic (because $\angle AEH + \angle AFH = 90^{\circ} + 90^{\circ} = 180^{\circ}$), $FECB$ is concyclic and $BD = FD = ED = CD$, we find $\angle HAF = \angle HEF = \angle BCF = \angle DFC = x$ and $\angle HAE = \angle HFE = \angle CBE = \angle DEB = y$. Then we find $\angle FED = \angle EFD = x + y$, and we know that $\angle HAF + \angle HAE = \angle BAC = x + y$, therefore $\angle FED = \angle EFD = \angle BAC$. Since we know that $\angle FED = \angle BAC$ and $\angle EFD = \angle BAC$, we conclude that $DE$ and $DF$ are both tangents to the circumcircle of triangle $AEF$.