Four Books on Conics (Commandino 1566)

L E M M A VI.

LET there be two equiangular parallelograms a c and d f, of which angle b is equal to angle e. I say that as the rectangle a b c is to the rectangle d e f, so is the parallelogram a c to the parallelogram d f.

Two similar parallelograms labeled a-b-g-c and d-e-h-f, with vertical perpendicular lines dropped from the top vertices to the bases.

For if the angles b and e are right angles, this is clearly evident: but if not, let perpendiculars a g and d h be dropped. And since angle b is equal to angle e; and the angle at g is a right angle equal to the right angle at h: the triangle a b g will be equiangular to the triangle d e h. Therefore,

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36. Book 1.

as b a is to a g, so is e d to d h. But as b a is to a g, so is the rectangle a b c to the rectangle contained by a g and b c: and as e d is to d h, so is the rectangle d e f to the rectangle contained by d h and e f. Therefore, by alternating, as the rectangle a b c is to the rectangle d e f, so is the rectangle contained by a g and b c; that is, the parallelogram a c, to the rectangle contained by d h and e f; that is, to the parallelogram d f.

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L E M M A VII.

LET there be a triangle a b c: and let b c be parallel to d e, and let the square which is made from c a be equal to the rectangle f a e. I say now if d c and b f are joined, the line b f is parallel to d c.

A triangle a-b-c with a parallel line d-e intersecting the sides. Lines connect d to c and b to f, where f extends beyond the base line.

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This, however, is clearly evident. For since as f a is to a c, so is c a to a e; and as c a is to a e, so is b a to a d: it will be as f a is to a c, so is b a to a d. Therefore, d c and b f are parallel to each other.

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L E M M A VIII.

LET there be a triangle a b c: and a trapezoid d e f g, such that angle a b c is equal to angle d e f. I say that as the rectangle a b c is to the rectangle which is contained by both d g, e f and d e, so is the triangle a b c to the trapezoid d e f g.

Three geometric figures: a triangle a-b-c with altitude a-h; a trapezoid d-e-f-g with altitude d-k; and a parallelogram d-e-f-g with altitude d-k.

FOR let perpendiculars a h and d k be drawn. And since angle a b c is equal to angle d e f; and the one at h is a right angle equal to the right angle at k; it will be as b a is to a h, so is e d to d k. But as b a is to a h, so is the rectangle a b c to that which is contained by a h and b c: and as e d is to d k...

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...to d k