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.