Mathematical Collections (Commandino 1588)

PROBLEM II. PROPOSITION VI.

It is required, given straight lines AB and BC, to find the mean in geometric analogy.

A geometric diagram shows a semicircle with diameter AB. A perpendicular line CD is drawn from the diameter to the circumference. Points A, D, and B are connected to form a triangle within the semicircle. Points E and F are marked on the diameter line.

Let CD be drawn at right angles, and let AB be bisected at point E. Let a circumference be described around the center E through B. Let it intersect the line drawn at right angles at D. From the point that joins BD, let an equal line BF be taken away; BF will be the mean we were seeking. For when DA is joined, it contains a right angle with BD, because each of them, BE and EA, is equal to that which A joins the points DE. Moreover, the angle at C is also a right angle; therefore, the triangle B ABD is equiangular to the triangle BCD. Because of this, the sides that are around their common angle B are proportional. Therefore, as AB is to BD, so is DB to BC; therefore, BD is the mean of the lines AB and BC, and BF is equal to it.

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A Because each of them, BE and EA, is equal to that which joins DE.] For ADB is the circumference of a semicircle, which comprehends a right angle, from the 33rd proposition of the third book of the Elements.

B Therefore, the triangle ABD is equiangular to the triangle BCD] from the eighth proposition of the sixth book of the Elements.

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PROBLEM III. PROPOSITION VII.

Given straight lines AB and BF, to take the minor extreme.

A geometric diagram similar to the one above features a semicircle on diameter AB with center E. A perpendicular line CD meets the circumference at D. A second arc is drawn from center F, intersecting the first at D. Points A, D, and B are connected.

Let AB be bisected at E, and let a circumference be described around center E through B; let it be intersected at point D by a circumference described around the center through F; and let the perpendicular DC be drawn. Thus, BC is the third proportional of the lines AB and BF, which can easily be demonstrated from the aforesaid concerning the mean proportional.

And it is manifest that if the given proportion of the analogy is double, such that AB is quadruple to BC, which is set equal to BD, then the half will be AB, namely EB. And if the proportion is greater than double, the half will be smaller; but if it is less than double, the half will be larger.