Taking my cue from "Trigonometry", by I. M. Gelʹfand, which left this proof as an exercise, I came up with the following:
Designate two Arbitrary Points "A" and "B" Fixed on a the same side of a diameter of a circle centered at point "O".
The arc between points A and B will subtend angle (AOB). For any point "C" on the designated circle outside the arc between A and B, the angle (ACB) will have exactly half the angular measure of (AOB), which must be constant for the fixed points "A" and "B". Also, the length of the secant [AB] will be constant, so that the ratio [AB] / Sin(ACB) will be constant, no matter where on the larger arc of the designated circle point "C" is located outside the lesser arc between A and B.
So for any triangle [ABC] so configured, the designated circle (centered at "O") will be its circumscribing circle (its 'circumcircle'), and the Law of Sines includes the ratios [BC] / Sin(BAC) = [AB] / Sin(ACB) for "a / Sin(A)" = "c / Sin(C)", so that we can use the special case of "C" being diametrically opposite point "B".
Then the angle (BAC) will have half the angular measure of the arc (BOC), the half-circle opposite point "A", making it a right angle, so that Sin(BAC) = 1.0, and [BC] will be a diameter of the circumcircle (with point "O" as its midpoint).
Then [BC] / Sin(BAC) = [DIAMETER] / (1.0) = [AB] / Sin(ACB) = the Circumcircle's Diameter = c / Sin(C) = b / Sin(B) = a / Sin(A).
Since "A" and "B" were fixed, but arbitrary, the previous sentence will be true for any triangle with its three points on the designated circle, and will extend to any circle which circumscribes any triangle.
(I only wish I had the wherewithal to include some fancy animations to illustrate the preceding proof.)