In isosceles $\triangle A B C$, it is known that $A B=A C$ $=k B C$, where $k$ is a natural number greater than 1. Points $D$ and $E$ are on $A B$ and $A C$ respectively, and $D B=B C=C E$. $C D$ and $B E$ intersect at $O$. 

Find the smallest natural number $k$ such that $\frac{O C}{B C}$ is a rational number.
II. Solution: As shown, connect $D E$, it is easy to know that $B C E D$ is an isosceles trapezoid. Also, from the given conditions, we have
$$
\angle 2=\angle 1=\angle 3 \text {, }
$$

thus $\triangle O B C \cos \triangle B C D$,
which means $O C \cdot C D=B C^{2}$.
Also, $\frac{C O}{O D}={ }_{D E}^{B C}=\frac{A B}{A D}$
$$
=\frac{A B}{A B-D B}=\stackrel{k}{k-1} \text {, }
$$
$\frac{C O}{C D}=\frac{k}{2 k-1}$.
(1) $\times$ ( 2 ), we get
$$
{ }_{B C} O C=\sqrt{\frac{k}{2 k-1}} .
$$

Since $\sqrt{\frac{k}{2 k-1}}$ is a rational number, and $k$ and $2 k-1$ are coprime, both $k$ and $2 k-1$ must be perfect squares.
$\cdots$ when $k=4,9,16$, $2 k-1=7,17$,

are not perfect squares.
When $k=25$, $2 k-1=49$ is a perfect square.
In summary, the smallest natural number $k$ that makes $\frac{O C}{B C}$ a rational number is $k=25$.