1. In $\triangle A B C$, it is known that $A B=A C, \angle C$'s bisector $C D$ intersects $A B$ at point $D, B D, B C, C D$ are three consecutive integers. Find the perimeter of $\triangle A B C$.
Let $A B=A C=b, B C=a$.
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
B D=\frac{a b}{a+b}, \\
B C-B D=a-\frac{a b}{a+b}=\frac{a^{2}}{a+b} \\
=k \in\{1,2\} .
\end{array}
$$

Following Example 1, we get
$$
C D^{2}=B D(B D+B C)=\frac{a^{2} b(a+2 b)}{(a+b)^{2}} \text {. }
$$

If $k=1$, then $b=a^{2}-a$.
Thus, $C D^{2}=(a-1)(2 a-1)>(a-2)^{2}$.
Therefore, $(a-1)(2 a-1)=(a+1)^{2}$.
Solving this, we get $a=5, b=20, a+2 b=45$.
If $k=2$, similarly, we get $a=3, b=\frac{3}{2}$, which is discarded.
Therefore, the perimeter of $\triangle A B C$ is 45.