Local Mathematical Competitions

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 $△CDE$ 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 $\angle BDA=\angle BXA=\angle CXE=\angle CDE$.

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Alternative solution.

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

Let $\gamma$ be the circumcircle of $△ABC$, and $\Gamma$ be the circumcircle of $△CDA$, 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 ($△CDA$) 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 $△CDA$.