Example 1 In $\triangle A B C$, $A B=A C, \angle A=$ $80^{\circ}, D$ is a point inside the shape, and $\angle D A B=\angle I B A=$ $10^{\circ}$. Find the degree measure of $\angle A C D$.

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Note: The symbol $\mathrm{I}$ in the original text seems to be a typo or a misprint, and it should likely be $D$ for consistency. If this is the case, the correct translation would be:

Example 1 In $\triangle A B C$, $A B=A C, \angle A=$ $80^{\circ}, D$ is a point inside the shape, and $\angle D A B=\angle D B A=$ $10^{\circ}$. Find the degree measure of $\angle A C D$.
Solution 1: As shown in Fig. $\mathrm{i}$, let $E$ be the point symmetric to $D$ with respect to the perpendicular bisector of $BC$. It is easy to see that $\triangle ADE$ is an equilateral triangle.

From the given information, we have $DA = DB$. Therefore, $EA = EC = ED$, which means $E$ is the circumcenter of $\triangle ADC$.
$$
\therefore \angle ACD = \frac{1}{2} \angle AED = 30^{\circ}.
$$

Here, an equilateral triangle is constructed, and a circumcenter is found, making the solution unconventional.

Solution 2: As shown in Fig. 2, construct an equilateral $\triangle ECA$ with $AC$ as a side inside $\triangle ABC$, and connect $BE$ and $DE$. It is known that $AD$ bisects $\angle BAE$. Therefore, $AD$ is the perpendicular bisector of $BE$. This gives $DE = DB$, and thus $DE = DA$. It follows that $CD$ is the perpendicular bisector of $AE$.
$$
\therefore \angle ACD = \frac{1}{2} \angle ACE = 30^{\circ}.
$$

Here, an equilateral triangle is constructed, and two easily provable facts about the concyclic points $A, F, E, C$ are used. It is known that $\angle DAF = 20^{\circ} = \angle DEF$, so $A, D, F, E$ are concyclic. Thus, $A, D, F, E, C$ are concyclic.
$$
\therefore \angle ACD = \angle AED = 30^{\circ}.
$$

Here, an equilateral triangle is constructed with $AB$ as one side, achieving the same goal in a different way. The construction is ingenious.

Solution 4: As shown in Fig. 4, construct an equilateral $\triangle EAB$ with $AB$ as a side inside $\triangle ABC$. Let the angle bisector of $\angle BAE$ intersect $BC$ at $F$, and connect $ED, EC, EF$. It is known that $\angle AFC = 80^{\circ}$. Additionally, $\angle EFC = \angle ABC = \angle ACB$, so $EC = ED = EF$. Therefore, $E$ is the circumcenter of $\triangle CDF$.
$$
\therefore \angle ACD = \frac{1}{2} \angle DEF = 30^{\circ}.
$$

A simple auxiliary circle is constructed, forming an equilateral triangle and six equal line segments, making the solution unique.

Solution 3: As shown in Fig. 3, with $D$ as the center and $DA$ as the radius, draw a circle intersecting $BC$ and $CA$ at $E$ and $F$ respectively. It is known that $B$ lies on the circle. Connect $DF, DE, EF$. It is easy to prove that $\triangle DEF$ is an equilateral triangle, and the perpendicular bisector provides convenience for solving the problem.