Coxeter- Exercise 2.5.5
- Pinyi Yao
Graph:first,three random points P,Q,R. Then a random point A on line RQ. Draw a line from A that is parallel to PQ, and this line intersects with RP at point B. Connect AP,QB,we get the intersection being point S. According to question and graph, PQ is parallel to AB, where A=PS·RQ, B=QS·RP. Angle QPS = Angle SAB, Angle PQS = Angle SBA. Triangle PQS and Triangle ABS are similar. Similarly, we can get that Triangle RPQ and Triangle RAB, Triangle GPS and Triangle CAS, Triangle RPG and Triangle RBC are similar. In this case, we know that GP/CA = PS/SA = PQ/AB = RP/RB = GP/CB. We get GP/CA = GP/CB. So CA=CB, in other words, C is the middle point of AB.