Example 6 In isosceles $\triangle A B C$, $\angle B=\angle C=40^{\circ}$, extend $A B$ to point $D$, such that $A D=B C$. Find the degree measure of $\angle B C D$.
Solution 1: As shown in Figure 7, construct an equilateral $\triangle ABE$ outside $\triangle ABC$ with $AB$ as a side, and connect $CE$. It is easy to see that
$$
AB=AC, \angle BAC=100^{\circ}.
$$

Since $\triangle ABE$ is an equilateral triangle, then
$$
\begin{array}{l}
AB=BE=AE, \\
\angle BAE=60^{\circ}, \\
\angle ABE=60^{\circ}.
\end{array}
$$

Therefore, $AC=AE$.
Thus, $\angle ACE=\frac{1}{2}(180^{\circ}-\angle CAE)=10^{\circ}$.
Hence, $\angle BCE=\angle ACB-\angle ACE=30^{\circ}$.
In $\triangle ADC$ and $\triangle BCE$, since
$AD=BC, AC=BE$,
$\angle DAC=\angle CBE=100^{\circ}$,
we have $\triangle ADC \cong \triangle BCE$.
Thus, $\angle ADC=\angle BCE=30^{\circ}$.
Therefore, $\angle BCD=\angle ABC-\angle D=10^{\circ}$.
Solution 2: As shown in Figure 8, construct an equilateral $\triangle BCE$ outside $\triangle ABC$ with $BC$ as a side, and connect $AE, DE$. It is easy to prove that
$\triangle ABE \cong \triangle ACE$.
$$
\begin{array}{l}
\text{Then } \angle AEC=\angle AEB=\frac{1}{2} \angle BEC=30^{\circ}, \\
\angle EAC=\angle EAB=\frac{1}{2} \angle BAC=50^{\circ}. \\
\text{Also, } \angle ACE=\angle ACB+\angle BCE \\
=40^{\circ}+60^{\circ}=100^{\circ}=\angle DAC, \\
AC=CA, CE=BC=AD,
\end{array}
$$

then $\triangle ACE \cong \triangle CAD$.
Thus, $\angle ACD=\angle EAC=50^{\circ}$.
Therefore, $\angle BCD=\angle ACD-\angle ACB=10^{\circ}$.
Solution 3: As shown in Figure 9, construct an equilateral $\triangle ADE$ with $AD$ as a side, such that point $E$ is inside $\angle CAD$, and connect $CE$. Then
$$
\begin{array}{l}
\angle CAE=\angle BAC-\angle BAE \\
=100^{\circ}-60^{\circ}=40^{\circ} \\
=\angle ABC.
\end{array}
$$

Also, $BC=AD=AE, AB=AC$, then
$\triangle ABC \cong \triangle CAE \Rightarrow AC=CE$.
Noting that $AD=DE$, we have
$\triangle ADC \cong \triangle EDC$.
Thus, $\angle ADC=\angle EDC=\frac{1}{2} \angle ADE=30^{\circ}$.
Therefore, $\angle BCD=\angle ABC-\angle ADC=10^{\circ}$.
Solution 4: As shown in Figure 10, construct an equilateral $\triangle ACE$ with $AC$ as a side, such that point $E$ is inside $\angle CAD$, and connect $DE$. Then
$$
\begin{array}{l}
\angle EAD=\angle BAC-\angle EAC \\
=100^{\circ}-60^{\circ}=40^{\circ}=\angle ABC.
\end{array}
$$

Also, $AE=AC=AB, AD=BC$, then
$\triangle EAD \cong \triangle ABC$.
Thus, $\angle AED=\angle BAC=100^{\circ}$,
$DE=AC=EC$.
Noting that
$$
\begin{array}{l}
\angle ECD=\angle EDC=\frac{1}{2}(180^{\circ}-\angle CED) \\
=\frac{1}{2}(180^{\circ}-100^{\circ}-60^{\circ})=10^{\circ}.
\end{array}
$$

Therefore, $\angle BCD=\angle ACE-\angle ACB-\angle ECD=10^{\circ}$.
Note: If there is an isosceles triangle in the problem with one of its internal angles being $20^{\circ}, 40^{\circ}$, or $100^{\circ}$, an equilateral triangle can be added to achieve the purpose of geometric quantity conversion. This is a commonly used method to solve relevant geometric problems.