Here is van Schooten's second problem and his first solution to it.
Problem II: Given a straight line with endpoints A, B, to bisect it.
Construction: Take any point C at random away from the line AB, and from A through C draw the indefinitely long line ACD, and locate on that line CD equal to the double of AC, and from D through B make a straight line. In that line, put BE equal to BD, and, joining C, E, I say it bisects AB at F.
Van Schooten then presented a second solution to Problem II.
Problem II: Given a straight line with endpoints A, B, to bisect it.
Another way: Find, as before, away from AB, any point C, and draw from it through the points A and B straight lines CAD, CBE, and make AD equal to AC and BE equal to BC. Then join DB, AE, and let them intersect at F. I say that if FC is made, it will bisect AB at G.
Finally, van Schooten presented a third solution to Problem II.
Problem II: Given a straight line with endpoints A, B, to bisect it.
Another way: Assume, as before, C is a point away from AB, and from B through C is drawn an indefinitely long line, and find on it CD equal to CB, and join it to A to form AD. Then in DA, assume DE equals DC and EF equals CB. [Here, van Schooten gives five illustrations, showing each of the following five cases: 1) DF = DA, 2) DF < DA, 3) DF > DA but DE < DA, 4) DE = DA, and 5) DE > DA. We illustrate case #2 only.)] Make BE, FC intersecting at G, then DGH cutting FB in H.
If now F falls on the point A, then the lines FB and AB coincide, and then H bisects it.
But if the point F falls beyond or within A, I say that if the line is drawn through the points C and H, it will bisect the line AB at I.