20.83 The sides of a cyclic quadrilateral are in sequence 25, 39, 52, and 60. The diameter of this circle is 
(A) 62.
(B) 63.
(C) 65.
(D) 66.
(E) 69.
(23rd American High School Mathematics Examination, 1972)
[Solution 1] Let the cyclic quadrilateral be $ABCD$. In $\triangle ABD$ and $\triangle CBD$, use the Law of Cosines, thus
$$
\begin{array}{l}
\overline{BD}^{2}=39^{2}+52^{2}-2 \cdot 39 \cdot 52 \cos C \\
\overline{BD}^{2}=25^{2}+60^{2}-2 \cdot 25 \cdot 60 \cos A
\end{array}
$$

Since $\angle C$ and $\angle A$ are supplementary, then $-\cos A=\cos C$. Substituting this into the equations and subtracting the two equations, we get
$$
0=0+(2 \cdot 39 \cdot 52+2 \cdot 25 \cdot 60) \cos A \text {, }
$$

Thus, $\cos A=0$. Therefore, $\angle A$ and its supplementary angle $\angle C$ are both right angles.
Then
$$
BD=\sqrt{39^{2}+52^{2}}=65 \text {. }
$$

Hence, the answer is $(C)$.
[Solution 2] Notice that $AB=25=5 \cdot 5, AD=60=12 \cdot 5$;
Also,
$$
BC=39=3 \cdot 13, \quad CD=52=4 \cdot 13 .
$$

Then
$$
\begin{array}{l}
AB^{2}+AD^{2}=5^{2}\left(5^{2}+12^{2}\right)=5^{2} \cdot 13^{2}, \\
BC^{2}+CD^{2}=13^{2}\left(3^{2}+4^{2}\right)=5^{2} \cdot 13^{2} .
\end{array}
$$

Thus,
$$
AB^{2}+AD^{2}=BC^{2}+CD^{2} \text {, }
$$

which means $\triangle ABD$ and $\triangle BCD$ are both right triangles, with $BD$ as the common hypotenuse.
Also, $2R=BD=\sqrt{5^{2} \cdot 13^{2}}=65$.
Hence, the answer is $(C)$.