Example 4 As shown in Figure $4, D$ is a point outside $\triangle A B C, A B=A C$ $=B D+C D, \angle A B D=$ $60^{\circ}$. Find the degree measure of $\angle A C D$.

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Solution 1: As shown in Figure 4, extend $BD$ to point $E$ such that $DE = CD$, and connect $AD$, $AE$.
From $AB = BD + CD = BD + DE = BE$,
$$
\angle ABE = 60^{\circ},
$$

we know that $\triangle ABE$ is an equilateral triangle. Therefore,
$\angle E = 60^{\circ}, AE = AB$.
Thus, $AC = AE$.
So, $\triangle ACD \cong \triangle AED$.
Hence, $\angle ACD = \angle E = 60^{\circ}$.

Solution 2: As shown in Figure 5, on $AB$, take $AE = CD$, and connect $DE$, and extend it to point $F$ such that $EF = AE$, and connect $AF$, $AD$. It is easy to see that
$$
\begin{array}{l}
BE = BD. \\
\text{Also, } \angle ABD = 60^{\circ},
\end{array}
$$

then $\triangle BDE$ is an equilateral triangle. Therefore,
$$
\angle BED = 60^{\circ}, BD = DE.
$$

Since $\angle AEF = \angle BED = 60^{\circ}, AE = EF$, thus, $\triangle AEF$ is an equilateral triangle.
Hence, $\angle F = 60^{\circ}, AF = AE$.
In $\triangle ACD$ and $\triangle DFA$, since
$AD = DA, CD = AE = AF$,
$AC = BD + CD = DE + EF = DF$,
therefore, $\triangle ACD \cong \triangle DFA$.
Thus, $\angle ACD = \angle F = 60^{\circ}$.