In $\triangle A B C$, $\angle B=50^{\circ}, \angle C=30^{\circ}, D$ is a point inside $\triangle A B C$, satisfying $\angle D B C=\angle D C B=20^{\circ}$. Find the degree measure of $\angle D A C$.
Solution: As shown in the figure, let $E$ be the circumcenter of $\triangle ABC$, and connect $AE$, $BE$, and $CE$. Since $\angle AEB = 2 \angle ACB = 60^{\circ}$, $\triangle ABE$ is an equilateral triangle, and $\angle ECB = \angle EBC = 10^{\circ}$, $\angle BEC = 160^{\circ}$.

On the extension of $BD$, take a point $F$ such that $CF = CB$, and connect $AF$ and $EF$.
$$
\because \angle BEC + \angle BFC = 160^{\circ} + 20^{\circ} = 180^{\circ},
$$
$\therefore B$, $E$, $C$, and $F$ are concyclic.
Thus, $\angle BFE = \angle BCE = 10^{\circ}$.
By the symmetry of the equilateral $\triangle ABE$,
$$
\angle AFB = \angle BFD = 10^{\circ}.
$$
$\vec{x} \angle DCA = 10^{\circ}$,
$$
\therefore \angle AFD = \angle DCA.
$$
$\therefore A$, $D$, $C$, and $F$ are concyclic.
Thus, $\angle DAC = \angle DFC = 20^{\circ}$.