1. The problem statement, all variables and given/known data a) Suppose that A,B,C,D are four "points" in a projective plane, no three of which are on a "line." Consider the "lines" AB, BC, CD, DA. Show that if AB and BC have a common point E, then E = B. b) From a) deduce that the three lines AB, BC, CD have no common point , and the same is true of any three of the lines AB, BC, CD, DA. 2. Relevant equations Axioms of a projective geometry: 1) Any two "points" are contained in a unique "line" 2) Any two "lines" contain a unique "point" 3) There are four different "points," no three of which are in a "line" 3. The attempt at a solution I proceeded by contradiction. Assume E is not equal to B. AB and BC have common point E by assumption so ABE are on a line and BCE are on a line. (I was tempted to say aha! contradiction - 3 points on a line right here, but it's not illegal, only illegal for the 4 points A,B,C,D to have 3 on line.) The E must connect to D by a line via Axoim 1. Either ED is along line DCE or DAE. If along DCE => B,C,D are along the same line if along DAE => D,A,B are along the same line. This contradictions that no 3 of the 4 points A,B,C,D are on a line. Therefore E must be equal to D. Don't think my argument is sensical. And, for part b, I'm confused since doesn't Axiom 2 state that any 2 lines must contain a point?