4. As shown in Figure 5, an equilateral $\triangle ABC$ with side length 26 is inscribed in a circle, and chord $DE \parallel BC$. The extension of $DE$ intersects the extensions of $AB$ and $AC$ at points $F$ and $G$, respectively. If the lengths of segments $AF$ and $DF$ are both positive integers, then the chord $DE=$ $\qquad$
4. 16 .

Let $A F=x, D F=y$. Then
$$
B F=x-26, D E=x-2 y, E F=x-y \text {. }
$$

By the secant theorem, we have
$$
\begin{array}{l}
A F \cdot B F=D F \cdot E F \\
\Rightarrow x(x-26)=y(x-y) \\
\Rightarrow x^{2}-(26+y) x+y^{2}=0 .
\end{array}
$$

Since $A F$ and $D F$ are both positive integers, we have
$$
\Delta=(26+y)^{2}-4 y^{2}>0 \Rightarrow y<26 \text {. }
$$

Upon verification, only $y=8$ satisfies the condition. Then $x=32$. Therefore, $D E=32-2 \times 8=16$.