Dividing a Semicircle into Proportional Parts
Proving the third of semicircle K o h lies between the parallels a b and c d
Leonardo poses the problem of cutting off any required fraction of a semicircle between two parallel lines. He divides the semicircle K o h into six equal sectors from the centre i, draws the parallels a b and c d, and counts the sectors and triangles that fall below c d. By adding the 'supplement' portion a o f b that lies above the parallel to the two triangles, he assembles four equal portions that together make up the third of the semicircle. A labelled semicircle divided into six sectors, cut by the parallels, illustrates the demonstration.
On this page
The proposition: cut any required part of a semicircle between parallels
The page opens with the general problem it sets out to solve: to give any part that may be required of a semicircle, contained between two parallel lines.
Six sectors, two parallels, and the counting of portions
The semicircle K o h is divided into 6 equal sectors K, c, a, o, b, d, h with their points at the centre i. Below the line c d remain 2 sectors c K i and d h i and 2 triangles c e i and e i d; adding the portion a o f b above (half for each triangle at the front e i) yields 4 portions that constitute the third of the whole.