Let point $X$ be lying on line $BE$ such that $\angle CXE = \angle CDE$ ($X$ is the (other than $C$) meeting point of the circumcircle of $\triangle CDE$ and the line $BE$).

Therefore the quadrilateral $DECX$ is cyclic, so $\angle DXE = \angle DCE = \pi - \angle CDA - \angle CAD$. But $\angle CDA = \frac{1}{2}(\pi - \angle ABC) = \angle CAB$, so $\angle DCE = \pi - \angle CAB - \angle CAD = \pi - \angle BAD$, hence the quadrilateral $ABXD$ is cyclic.

It follows $\angle BDA = \angle BXA = \angle CXE = \angle CDE$.

We are asked to prove that $DE$ and $DB$ are isogonal conjugate, and so, since $DE$ is median in $\triangle CDA$, that $DB$ is symmedian in that triangle.

Let $\gamma$ be the circumcircle of $\triangle ABC$, and $\Gamma$ be the circumcircle of $\triangle CDA$, of center $\Omega$. We have $\angle C\Omega A = 2\angle CDA = \pi - \angle ABC$, hence $\Omega \in \gamma$, and $C\Omega = A\Omega$, hence $\Omega \in BE$; therefore $\Omega = \gamma \cap BE$. Since $B\Omega$ is a diameter for circle $\gamma$, it follows $\angle B\Omega A = \angle BC\Omega = \pi/2$, therefore $BA$ and $BC$ are tangent to circle $\Gamma$.

A well-known LEMMA states that a symmedian in a triangle ($\triangle CDA$) connects the vertex ($D$) it originates at with the intersection ($B$) of the tangents ($BA$ and $BC$) to the circumcircle ($\Gamma$) of the triangle at the other two vertices ($A$ and $C$); for us that yields $DB$ symmedian in $\triangle CDA$.