Example 7 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=30^{\circ}, R$ is a point inside the triangle, $\angle R B C=$ $\angle R C B=20^{\circ}$. Find the degree measure of $\angle R A B$.
Solution: As shown in Figure 7, construct a regular $\triangle DAB$ on one side of $AB$ inside $\triangle ABC$. Connect $DR$ and $DC$.

Given $\angle ACB = 30^{\circ}$, we know that point $D$ is the circumcenter of $\triangle ABC$. Therefore, $DB = DC$. Thus, $\angle DCB = \angle DBC = 10^{\circ}$, and $\angle BDC = 160^{\circ}$.

Given $\angle RBC = \angle RCB = 20^{\circ}$, we know that $RB = RC$. Therefore, $RD$ is the perpendicular bisector of $BC$, and
$$
\angle RDB = \frac{1}{2} \angle BDC = 80^{\circ}.
$$

Given $\angle RBA = 30^{\circ}$, we know that point $A$ is symmetric to point $D$ with respect to $BR$. Therefore, $\angle RAB = \angle RDB = 80^{\circ}$.

Constructing a regular $\triangle ABD$ on one side of $AB$ inside $\triangle ABC$ essentially finds the symmetric point of $A$ with respect to $BR$. Since point $D$ lies on the perpendicular bisector of $BC$, it makes the solution very convenient.