Constructing a line equal to a segment's remainder
Two intersecting circles cut an oblique equal to the leftover part of a line
This verso sets out a geometric construction: from a given point on a straight line, to draw an oblique line whose length equals the remaining part of the line it crosses. Leonardo solves it by drawing two equal circles centred on the given point and on the line's end, joining their two intersections, and taking where that join cuts the base line. Diagrams at right show the intersecting circles lettered a, b, c, n, m, g, together with a radiating arc at the top, and the reasoning rests throughout on the definition of the circle.
On this page
Drawing an oblique equal to the hidden part of a line
From a given point on a straight line, Leonardo seeks an oblique line ending on the same straight line and equal to the portion of it that lies hidden beneath the oblique. He makes two circles of equal diameter, one centred on the given point and one on the line's end, and joins their two points of intersection to reach the sought endpoint.
Restatement with the point a on line f c
The problem is restated for the point a on the line f c: to cut f e with an oblique a b m so that the remainder b c equals the oblique a b. Leonardo directs the reader to the worked solution set out below the figure.
Solution by two intersecting circles
Taking the given point and the end c as centres of two circles, their intersections define the line n m, which cuts f e at b so that a b equals b c by the definition of the circle. The circle a c e cuts the indefinite line at g, giving the length into which c a is transmuted as c g while keeping the same measure.