6. As shown in Figure 2, given a convex quadrilateral $ABCD$ satisfying $\angle CAD=45^{\circ}$, $\angle ACD=30^{\circ}$, $\angle BAC$ $=\angle BCA=15^{\circ}$. Then the degree measure of $\angle DBC$ is
$6.90^{\circ}$.
As shown in Figure 5, let $K$ be a point on $CD$ such that $\angle CAK = 30^{\circ}$. Then $BK$ is the perpendicular bisector of $AC$, and
$$
\begin{array}{l}
\angle AKB \\
=\angle BKC \\
=60^{\circ}.
\end{array}
$$

Thus,
$$
\angle AKD = 60^{\circ}.
$$

Since $\angle BAD$
$$
= 60^{\circ}, \text{ therefore, } A, B, 
$$
$K, D$ are concyclic.
Slightly 5

Hence, $\angle DBK = \angle DAK = 15^{\circ}$.
Also, $\angle KBC = \angle DBC - \angle DBK = 75^{\circ}$, so
$$
\angle DBC = \angle DBK + \angle KBC = 90^{\circ}.
$$