Squaring the circle: circles, bi-angles and curvilinear triangles
Geometric demonstrations equating the areas of circles, squares and bi-angular figures
A geometry sheet from Leonardo's studies on the quadrature of the circle, written in his usual mirror hand at the upper left and upper right. The right margin carries a stack of diagrams: circles enclosing two smaller circles and curvilinear triangles, a square framing inscribed circles, and a square with an inscribed circle and rotated square, with two further circle figures at lower right. The text argues by equal areas that two circles equal half the greatest circle, that the remaining curvilinear triangles equal the other half, and that combining the first and second figures yields a square. A later foliation numeral is also visible on the sheet.
On this page
Two circles worth half the greatest circle; the curvilinear triangles the rest
The two circles a b are worth half of the greatest circle, so the two curvilinear triangles are worth the other half. Removing the two bi-angles n p is like removing the four greatest portions from a circle equal to half the greatest circle. What then remains, not hatched, equals half of the greatest square of the greatest circle.
First and second figures joined make a square
The square's empty, hatched part equals the un-hatched figure above it. Therefore the first and second figures joined together are worth the third, which is a square. Squaring the first thus gives the smallest square of that third figure below.
Beginning of the third demonstration
Marked 'Third, o a', the note opens a further demonstration that continues off the transcribed passage. It is keyed to one of the lower diagrams on the sheet.