According to the figure, using the idea in exercise 2.1.3, we are calling the circle through the three diagonal points (A,B,G) a line, then A,B,G are collinear. We know that G=PQ·RS. Since we are construcing a line AB to form the harmonic set, and we also know that G is on line AB and line RS. So G=AB·RS=C. Because G is also on PQ, F=PQ·AB=G. In this case, G,C,F are the same point. That means there are three points in the harmonic set, being A,B,G(or C, or F).