Triangles between parallels and circular portions
Curved and straight triangles on the same base proved equal, with eight circles in eighths
Continuing the quadrature studies, Leonardo proves that triangular figures — straight-sided or curved ('orbicular') — standing on the same base between parallel lines are all equal in area. He supports this with a struck-through 'orbicular pyramid' and a row of triangles between parallels, and states that of segments on an equal chord the one with the longer sagitta has the greater capacity. A row of eight numbered circles divided into eighths, and a note that a seventh of one number can equal an eighth of another, extend the reasoning to proportions between circles.
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Triangles on the same base between parallels are equal
Between two parallel lines Leonardo draws several triangles on an equal base, straight-sided and curved, and proves them all equal in area because each equals the triangle a b c. He extends the argument from rectilinear to curvilinear 'parallel spaces', treating the circle's circumference as a line parallel to itself about the center.
Orbicular triangles equal to straight ones
A struck-through figure of an 'orbicular pyramid' accompanies the claim that all triangular surfaces, straight or curved and however tall, built on the same base with uniformly tapering sides are equal to one another.
Segment capacity grows with the sagitta
A marginal rule states that among circular segments having an equal chord, the one with the longer sagitta — the more curved arc — has the greater capacity.
Sevenths and eighths of two circles
A row of eight numbered circles, each taken in eighths, leads to the rule that a seventh part of one number can equal an eighth part of another. Taking 8 and 7, Leonardo shows the unit composing 8 worth the unit composing 7, so that of two circles in the ratio 7/8 the units removed are equal.
