1972 AMC 12 Problem 25

Notice two scaled Pythagorean triples among the four side lengths

The $25,60$ pair and the $39,52$ pair share the same possible hypotenuse

The consecutive sides may be labeled $$\begin{gathered} AB=25,\\ BC=39,\\ CD=52,\\ DA=60. \end{gathered}$$ Since $$25^2+60^2=65^2=39^2+52^2,$$ triangles $ABD$ and $BCD$ are right triangles with common hypotenuse $BD=65.$ Their union is the cyclic quadrilateral, and a right triangle’s hypotenuse is a diameter of its circumcircle. Thus the diameter is $65.$

Therefore, the correct answer is C.