Problem 369

4. (D).

Let $A B C D$ be a cyclic quadrilateral, and $A B=25$,
$$B C=39, C D=52, D A$$
$=60$. By the property of a cyclic quadrilateral, $\angle A=180^{\circ}-\angle C$. Connecting $B D$. By the cosine rule,
$$\begin{aligned} B D^{2} & =A B^{2}+A D^{2}-2 A B \cdot A D \cos \angle A \\ & =C B^{2}+C D^{2}-2 C B \cdot C D \cos \angle C, \end{aligned}$$

which means
$$\begin{array}{l} 25^{2}+60^{2}-2 \cdot 25 \cdot 60 \cdot \cos \angle A \\ =39^{2}+52^{2}+2 \cdot 39 \cdot 52 \cdot \cos \angle A . \end{array}$$

Therefore, $\angle A=90^{\circ}, B D$ is the diameter of the circle.
$$B D=\sqrt{25^{2}+60^{2}}=\sqrt{4225}=65 \text {. }$$

Thus, the correct choice is (D).