What you're asking is whether the sums of the sides of all triangles inscribed in a circle, divided by 3, are the same. That's the same as saying that the sums of the sides of all such triangles are the same, WITHOUT dividing by 3. That would have to be the case if what you proposed was true.
But that's not the case.
Let's start with an equilateral triangle. I know that the length of each side of an "inscribed" equilateral triangle is √3 times the radius of the circle, so the sum of the sides is 3√3 r.
To see that this is not the same as the sum of the sides of any inscribed triangle, draw a triangle like this one: Start with a point close to the top of the circle (from the point of view of the paper you drew it on.) Now draw one side that is ALMOST a diameter of the circle. It passes just slightly to the left of the center and crosses to a point almost exactly opposite the first point. Now draw another side that also starts from the top point but passes just to the RIGHT of the center, and goes to another point almost exactly opposite the first point. And finally, the third side of the triangle connects the two opposite points you created, which are actually very close to each other.
Now you have a triangle with two long sides, almost equal to the diameter of the circle, and one short side, almost equal to 0. The closer you make the two points opposite the original point, in fact, the closer the long sides are to the diameter (2r) and the closer the short side is to 0.
So if you make the two bottom points VERY VERY close to each other, the lengths of the two long sides add up VERY VERY close to d + d = 2r + 2r = 4r, and the length of the short side is VERY VERY close to 0. If you add those up, you get a total of the sides that is VERY VERY close to 4r. That can't be the same as 3√3 r. So that shows that the sum of the sides of any random inscribed triangle is not necessarily going to be the same as the sum of the sides of an inscribed equilateral triangle.
And there are other ways to show this, using other example "random" triangles.
So your procedure is not going to work.