Answer:
It isn't.
The correct relation is 4r·Area = abc.
Explanation:
You want to show the relation between the product of triangle side lengths and the product of the triangle area and the radius of the circumcircle.
Triangle area
The area of the triangle in the attached diagram can be found by ...
Area = 1/2(bc)sin(α)
Side length
The inscribed angle theorem tells you the measure of the arc subtended by an inscribed angle is twice the measure of the inscribed angle. That means the opposite chord will be twice the length of the product of the radius and the inscribed angle value:
a = 2r·sin(α)
Circumradius product
The product of the circumradius and the area of the triangle is ...
r·Area = r·(1/2)(bc)sin(α) = 1/4(bc)(2r·sin(α)) = 1/4abc
Without the fractions, the relation is ...
4r·Area = abc
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Additional comment
As a simple example, consider a 3-4-5 right triangle. The area of the triangle is ...
A = 1/2(3)(4) = 6
The circumradius is half the length of the hypotenuse, r = 5/2.
The relation of interest is ...
4r·Area = abc
4(5/2)(6) = 3·4·5
60 = 60 . . . . . . . . . . as expected
This could be a very simple way to find the circumradius of a triangle.