In this article, we mean to discuss the topics from Book II of van Schooten's Exercitationum mathematicarum libri quinque, but first it would be a good idea to look at the book as a whole.
Book I is a fairly standard review of arithmetic and ordinary geometry.
Book II contains the ruler constructions that we will get to later.
In Book III, van Schooten tries to reconstruct some of the works of Apollonius on plane loci. This was an important research topic of the time, and both Fermat and Descartes devoted a lot of effort to it as well. As usual, van Schooten is about 15 years late, though.
Ellipse drawer from Book IV of van Schooten's Exercitationes mathematicae
Book IV contains van Schooten’s best known work. Its title is “Organica conicarum sectionum”, or “Instruments of conic sections.” The word “organica” is more closely related to the organ as a musical instrument than to the “organic” we sometimes encounter in chemistry or agriculture. As the title suggests, the chapter describes a variety of beautiful instruments for drawing the various conic sections. One of several mechanisms for drawing a parabola is shown on the preceding page of this article, and another for drawing an ellipse is pictured here. All of the instruments are shown being operated by elegant disembodied hands, sometimes showing the cuffs of beautifully ruffled sleeves fashionable at the time. Several of the illustrations include beautiful and irrelevant doodles, like the rosette beside the ellipse machine. For animated versions of van Schooten's parabola and ellipse drawers, see Figures 2, 6, and 8 in the module "Curve Drawing Then and Now" in the Convergence article, "Historical Activities for the Calculus Classroom."
Finally, Book V is titled “Sectiones triginta miscellaneas”, or Thirty miscellaneous sections. In these sections, van Schooten carefully develops the combinatorial principles of counting. It is probably his best work, and it provides a complete foundation for the work of Huygens that follows in the appendix.
Then, van Schooten uses these new counting principles to rediscover the two new pairs of amicable numbers that Fermat and Descartes had discovered fifteen years earlier. Remember that a pair of numbers is amicable if the proper factors of one number add up to the other, and vice versa. For 2000 years, 284 and 220 were the only known pair of amicable numbers. Just a couple of years apart, Fermat and Descartes each discovered another pair. In fact, those two pairs had been discovered a couple of centuries earlier by Arab mathematicians. Van Schooten, of course, knew about the pairs that Fermat and Descartes had discovered, but his approach showed how both pairs could be discovered using the same method. It gave hope that more pairs could be discovered in the same way. Alas, the technique was a dead end, and no new amicable pairs would be discovered for another hundred years.
We return to Book II, “Constructio problematium simplicium geometricorum”, Solution of simple problems of geometry. Van Schooten gives us three postulates. One may
We might call these “ruler constructions”, because we are not allowed to use a compass in the geometric constructions. It might be more rigorous to call them “straightedge constructions,” because we are not allowed to use the instrument to measure anything.
Postulate 3 makes van Schooten’s ruler more powerful than Euclid’s ruler was. We are allowed to use our ruler to copy the length of one line segment onto another line. The postulate does not tell us how to do this, only that it is possible to do it. In practice, we can copy a segment by making a mark on the ruler.
Van Schooten gives us his ruler constructions as a series of ten problems, with solutions. For some problems, he gives two or more solutions. He begins with bisecting a given angle.
Problem I. Given a rectilinear angle BAC, to cut it into two equal pieces.
Details of first solution to Problem I
Quicker version of first solution to Problem I (verbatim translation of van Schooten’s solution)
We repeat Problem I, trisecting a given angle.
Problem I. Given a rectilinear angle BAC, to cut it into two equal pieces.
The details of van Schooten's first solution to this problem are as follows.
Given the angle BAC, 

copy the segment AC onto the line AB to get the segment AD equal to AC. 

Likewise, copy the length AB onto the line AC, extended if necessary, to get the segment AF equal to AB. 

Now, draw the straight lines BC and DF, and let them intersect at the point G. 

Now, the line AG will cut the angle into two equal pieces. 
The sharp reader will notice that in this solution, we seem to assume that segment AC is shorter than segment AB. Though that can easily be circumvented, the solution will not quite work as given in the special case that AC equals AB. These issues do not seem to bother van Schooten at all.
We repeat Problem I, trisecting a given angle.
Problem I. Given a rectilinear angle BAC, to cut it into two equal pieces.
In the detailed version of the solution to Problem I given on the preceding page, we slightly misrepresented the given data. In fact, when van Schooten gives us the angle BAC, he seems to be assuming that AB is a segment, and that C is a segment with A as one of its endpoints. It could also be that B and C are both line segments, sharing A as an endpoint, and that the symbol B is also used to denote a point on the segment B. This seems to have no real consequence, but it is curious. Van Schooten’s words translate as follows.
Suppose that in the line AB points B and D are placed. and are placed on line AC so that AE equals AD, and E is a point further along AC so that EF equals DB; And draw straight lines BE, DF, which intersect at G. I say that the line AG cuts the angle BAC into two equal pieces.
Van Schooten also includes a proof of the correctness of each of his constructions, but we will generally leave those to the reader.
Van Schooten offered a second solution to the problem of bisecting an angle BAC. He wrote:
Another way: Assuming the point B is in AB, find in CA, produced through A, the segments AD, DE, each equal to AB, and draw the straight line through the points E, B, and put on it BF equal to BE, and join AF. I say it bisects that angle BAC.
The proof of correctness of this construction may not be immediately obvious. As a hint, we suggest that a line from B to D might be useful.
We repeat Problem I, trisecting a given angle.
Problem I. Given a rectilinear angle BAC, to cut it into two equal pieces.
Van Schooten wrote a third solution.
And another way. Assume, as before, AD equals AB, that BD are joined, and that BE equals BA. And from E through D an indefinitely long straight line is drawn. In that, if DF is put, equal to DE, and from F through H [sic: should be A] is drawn FA: I say that this bisects the angle BAC.
We repeat Problem I, trisecting a given angle.
Problem I. Given a rectilinear angle BAC, to cut it into two equal pieces.
Van Schooten wrote a fourth solution.
Again another way: Assume in AB the given point G, and from A is located in AC a line AF equal to AB; CA is produced to E so that EA is either larger or smaller than AB. Then produce BA from the point A and place a straight line AD equal to AE; and make through the points E, B and D, F straight lines EBG and DFG, which intersect at G. This done, if the straight line is drawn through AG, it will bisect that angle BAC.
Van Schooten gives two diagrams, depending on whether EA is larger or smaller than AB. We give here the illustration taking EA larger than AB.