The geometrical proof proceeds thus. Drawing in the segment the inscribed triangle with the same base and height as the segment, Archimedes next inscribes triangles in precisely the same way in each of the segments left over, and proves that the sum of the two new triangles is ¼ of the original inscribed triangle. Again, drawing triangles inscribed in the same way in the four segments left over, he proves that their sum is ¼ of the sum of the preceding pair of triangles and therefore (¼)² of the original inscribed triangle. Proceeding thus, we have a series of areas exhausting the parabolic segment. Their sum, if we denote the first inscribed triangle by Δ, is
Δ {1 + ¼ + (¼)² + (¼)³ + . . . .}
Archimedes proves geometrically in Prop. 23 that the sum of this infinite series is 4⁄3Δ, and then confirms by reductio ad absurdum the equality of the area of the parabolic segment to this area.
CHAPTER V.
THE SANDRECKONER.