Fortification or Military Architecture

Signature: A 3

equal parts; The reason is that it is already halved at E, but EH is also halved at point 1, for from centers H and E, arcs have been made that intersect at F and G, consequently, etc.

Figure 47.

To divide the line AB into as many equal parts as one wishes, as for example into 5.

Having made two parallel lines AD and BC as one wishes upon the extremities, then taken upon each 4 parts of the same opening original: "ouverture"; this refers to keeping the compass at a fixed width for each measurement, the lines that join them will divide the proposed line into 5 equal parts; one must always take one part fewer on the parallels Marolois is explaining that to divide a line into n segments, you only need to mark n-1 points on the construction lines..

Figure 48.

Through 3 points A, B, and C, which are not in a straight line, to pass the circumference of a circle.

One shall cut the imaginary line AB into two equal parts by a perpendicular FG according to the Corollary of Figure 44; likewise the imaginary line BC by DE, which will intersect at the center of the required circle K; and having taken an interval original: "interval"; in 17th-century geometry, this term refers to the radius or the distance between the two points of the compass to one of the given points, one shall make a circle which will pass through the said 3 points.

Figure 49.

To make an arc parallel to a given arc ABC, whose center is unknown.

After having made 3 points A, B, and C upon it and found the center K by the preceding [figure], one shall make a circle of such interval as one wishes upon the said center K.

Figure 50.

Through a point B, to trace a parallel to the given line CD.

Upon [the point] as a center, let an arc be made touching the line at D; and with the same interval, from center C wherever one wishes on CD, let a similar [arc] be made reciprocally; then, drawing BA touching the last arc, it will be parallel to CD.

Otherwise Figure 51.

Through a point C to lead a parallel line to AB.

Having taken the distance AC and placed it at BD, let then from the center and of the interval AB an arc be made cutting the other at D; then CD will be parallel to AB.