1. **Problem Statement:** (i) Given four points A, B, C, D in a projective plane, no three collinear, consider lines AB, BC, CD, DA. Show that if lines AB and BC have a common point E, then E = B. 2. **Key Concept:** In a projective plane, two distinct lines intersect in exactly one point. Since AB and BC share point B, if they have another common point E, then E must be B because no two distinct lines can intersect in more than one point. 3. **Proof for (i):** - Lines AB and BC both contain point B. - Suppose they have a common point E. - Since no three points are collinear, E cannot be any other point on AB or BC except B. - Therefore, E = B. 4. **Deduction for (ii):** - From (i), lines AB and BC intersect only at B. - Consider three lines AB, BC, CD. - If they had a common point, it would have to lie on AB and BC, which is B. - But CD does not pass through B (no three points collinear), so no common point exists for all three lines. - Similarly, for lines AB, BC, CD, DA, no three share a common point. 5. **Example for (iii):** - In the real projective plane \(\mathbb{R}P^2\), consider three lines: - \(l_1: x=0\) - \(l_2: y=0\) - \(l_3: x+y+z=0\) (in homogeneous coordinates \([x:y:z]\)) - These three lines intersect pairwise but no single point lies on all three. **Final answers:** (i) If AB and BC have a common point E, then \(E = B\). (ii) The three lines AB, BC, CD have no common point, and the same holds for any three lines among AB, BC, CD, DA. (iii) Example lines in \(\mathbb{R}P^2\) with no three concurrent points are \(x=0\), \(y=0\), and \(x+y+z=0\).