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Denote $\angle BDC=x$ and $\angle DBA=y$. Then $\angle BAC=2x$. Let $P$ and $Q$ be the intersections of the angle bisector of $\angle BAC$ with segment $\overline{BC}$ and ray $DC$.



Since $\angle BAQ=\angle BDQ=x$, the quadrilateral $ABQD$ is cyclic. Given that $\angle BAD$ is a right angle, $\angle DQB$ is a right angle as well.

Inscribed angles $\angle DQA$ and $\angle DBA$ subtending the $\overline{AD}$ are equal, i.e. $\angle DQA=y$, while $\angle AQB=90^{\circ }-y$.

From $\angle BCQ=180^{\circ }-\angle DCB=y=\angle DQA$, we see that the triangle $PQC$ is isosceles and that $|PQ|=|PC|$.

Since $\angle QBP=\angle QPC-\angle PQB=180^{\circ }-2y-(90^{\circ }-y)=90^{\circ }-y=\angle AQB$, triangle $PBQ$ is isosceles as well and $|PB|=|PQ|$.

Hence, $P$ is the midpoint of $\overline{BC}$ and, by the converse of the angle bisector theorem applied to $\angle BAC$, we conclude that $|AB|=|AC|$. Therefore $90^{\circ }=\angle APB=\angle QPC=180^{\circ }-2y$, which implies that $\angle DBA=y=45^{\circ }$.