Inscribing a hexagon and pentagon; area of a quarter-corona
A circle holds five triangle-bases; comparing hexagon and pentagon radii
Leonardo compares the areas and inscribed polygons of circles. He argues that a quarter of a circular corona equals three fourths of the circle below it, entering that circle one and one-third times, and that a circle centred on the middle of an equilateral triangle's base and passing through its apex contains five of that triangle's bases along its circumference. Superposed circles carry an inscribed hexagon (labelled b, h, g, f, e, d, i) and pentagon (a, b, c, k, n, m), used to prove the smaller circle is one-sixth less than the larger. The construction is repeated at the left of the page.
On this page
The quarter-corona and the circle beneath (a b c d)
The quarter-corona a b c d enters one and one-third times into the circle underneath, so it holds three fourths of that circle's area. The proof is set out at the right, below, in square a b h r.
Hexagon and pentagon inscribed in superposed circles
Two superposed circles carry an inscribed hexagon and pentagon, their radii dividing each polygon into equal triangles. Vertices of the hexagon are labelled h, g, i, b, f, d, e; those of the pentagon a, c, k, n, m, with the uppermost hexagon triangle marked false.
A circle containing five bases of the triangle
A circle whose centre sits over the middle of an equilateral triangle's base and whose circumference passes through the triangle's apex must contain five of that triangle's bases along its line. The claim is stated as a proposition.
Proof: the smaller circle is one-sixth less
Since every radius is equal, lines b e and b d equal d e and complete the equilateral triangle, six of which make the hexagon d e f g h i. Line b a equals b d and, set five times into circle a b k n m, composes a regular pentagon; the smaller circle comes out one-sixth less than the larger, its line being five lengths against the larger's six.