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 $\angle DBA = y = 45^\circ$.