Trisecting a Circular Portion, Concluded
Finishing the centre of accidental gravity and dividing a portion into three equal parts
The page opens by completing the previous statement — the centre of accidental gravity lies at a quarter of the length toward the base — then turns to dividing a circular portion into three equal parts. Assigning triangle b e f to triangle a b e gives part a b f as one third, and turning triangle b g h to g h c gives part g c d as a second third, leaving b c f g as the final third. Leonardo declares the demonstration concluded.
On this page
The centre of accidental gravity at a quarter of the length
Completing the preceding folio, the centre of accidental gravity is placed at the fourth of the body's length toward the base.
Assembling the three equal thirds of the portion
Leave triangle b e f to triangle a b e so that part a b f is one third of the portion; turn triangle b g h to g h c so that part g c d is a second third; the remaining part b c f g stands for the last third, and the intent is concluded.