5. As shown in Figure 6, in isosceles $\triangle ABC$, it is known that $AB = AC = k BC$ ($k$ is a natural number greater than 1), points $D$ and $E$ are on sides $AB$ and $AC$ respectively, and $DB = BC = CE$, $CD$ intersects $BE$ at point $O$. Find the smallest positive integer $k$ such that $\frac{OC}{BC}$ is a rational number.
(1991, Shanghai Junior High School Mathematics Competition)
Connect $D E$. It is easy to see that quadrilateral $B C E D$ is an isosceles trapezoid.
By the given condition $\angle B D C=\angle B C D=\angle E B C$, hence
$$
O C \cdot C D=B C^{2} \text {. }
$$

On the other hand,
$$
\frac{O C}{O D}=\frac{B C}{D E}=\frac{A B}{A D}=\frac{A B}{A B-D B}=\frac{k}{k-1} .
$$

Therefore, $\frac{O C}{C D}=\frac{k}{2 k-1}$.
(1) $\times$ (2) gives $\frac{O C}{B C}=\sqrt{\frac{k}{2 k-1}}$ is a rational number, and $(k, 2 k-1)=1$.
Thus, both $k$ and $2 k-1$ are perfect squares.
When $k=4,9,16$, $2 k-1=7,17,31$;
When $k=25$, $2 k-1=49$.
Therefore, the smallest positive integer $k$ that makes $\frac{O C}{B C}$ a rational number is $k=25$.