$20 \cdot 84$ If the sides of a cyclic quadrilateral are successively $25, 39, 52$, and 60, then the circumference of the circle is
(A) $62 \pi$.
(B) $63 \pi$.
(C) $64 \pi$.
(D) $65 \pi$.
(China Junior High School Mathematics League, 1995)
[Solution] Let $ABCD$ be a cyclic quadrilateral, and $AB=25, BC=39, CD=52, DA=60$. By the property of a cyclic quadrilateral, $\angle A = 180^{\circ} - \angle C$.
Connect $BD$. By the Law of Cosines,
$$
\begin{aligned}
BD^2 & = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos \angle A \\
& = CB^2 + CD^2 - 2 \cdot CB \cdot CD \cdot \cos \angle C,
\end{aligned}
$$

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

Solving for $\cos \angle A = \frac{25^2 + 60^2 - 39^2 - 52^2}{2(25 \cdot 60 + 39 \cdot 52)} = 0$.
Thus, $\angle A = 90^{\circ}$, and $BD$ is the diameter of the circle.
$\therefore BD = \sqrt{25^2 + 60^2} = \sqrt{4225} = 65$. The circumference of the circle is $65\pi$. Therefore, the answer is $(D)$.