Fortification or Military Architecture

The Propositions Follow.

Plate 4. Figure 43

To describe an equilateral triangle on a straight line A B.

From Center A & Distance original: Interval, referring to the distance between the two points of a compass A B let the arc BC be made, cutting the arc of the same Distance, & from Center B at point C, from which drawing the straight lines C A, CB, the triangle A C B will be equilateral: For CB is equal to B A; and C A to A B by the definition of the Circle.

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Figure 44.

To divide the line A B into two equal parts.

From Center A & with a distance greater than half of A B let the arc C D be made, and from Center B, with the same distance let another arc be made cutting the first at points C, D, through which passing a line C D, it will cut A B through the middle E.

Corollary.

It appears from this how one shall cut a line A B in the middle, and at right angles by another CD, for if one draws CA, AD, DB, B C it will be proved that the 4 angles at point E are right angles.

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Figure 45.

To cut a line into three equal parts, using a single opening of the compass.

This construction is part of a tradition of "fixed-compass" geometry, where the geometer attempts to solve problems without changing the width of the compass after the first measurement.

Let A B be the given line, from the distance of which, let arcs be made, from centers A, B; by the preceding [proposition] AB & CD are mutually divided into two equal parts at P: from the same distance, and center D let there be a Circle, then FH, EG of the same opening, and drawing HC, CG, which will divide the line AB into 3 equal parts; since AB is equal to H G, then D will be in the middle of P O, and thus will make C O in 3 equal parts, which are proportional to the parts of AB, for as O C to CP is 3 to 1 so HG (or AB) to the middle part which will then be the third of AB, etc.

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Figure 46.

To cut B H into 4 equal parts, using a single opening of the Compass.

Having drawn CD from the intersections of the arcs described from centers B, H, then a Circle from Center E (all with the distance of the line BH) cutting the aforementioned arcs at 4 points, from which the lines F G, IK drawn, will cut HB into 4

A 2