A new area-maximization proof for the circle

Conclusion  Finally, we compare the trianglesT _i with the isosceles triangles which would be obtained by slicing the circle by the same procedure. Since they all have the same short base length and the same angle opposite the base, the isosceles triangles composing the circle have more area than the trianglesT _i. Because the latter triangles cover all ofR ₁ (perhaps even with overlap and/or extension beyondR ₁), we see that the area of the quarter-circle is greater than that ofR ₁, and thus the circle's entire area is greater than that of our original regionR.