A Finite Projective Geometry Exercise 1.16.
Do not worry about your difficulties in mathematics; I assure you that mine are greater.
—Albert Einstein (1879–1955)

Exercise 1.16.  Prove the Dual of Axiom P3. Every point has at least three lines incident with it.

Proof. Let A be a point. By the Lemma below, there are points B, C, and D such that no three of A, B, C, and D are collinear. By Axiom P1, there are lines AB, AC, and AD. (We need to show the lines are distinct.) Since no three of A, B, C, and D are collinear, C and D cannot be on line AB. Similarly, B and D cannot be on line AC, and B and C cannot be on line AD. Hence AB, AC, and AD are three distinct lines incident with A. Since A was arbitrarily chosen, every point has at least three lines incident with it.//

Lemma. For every point A there are three distinct points B, C, and D such that no three of A, B, C, and D are collinear.

Proof. Let A be a point. Suppose the Lemma is not true. Then for any set of four points that include A, three of the points are collinear. By Axiom P4, there exist at least four points B, C, D, and E such that no three of them are collinear. Note that A cannot any of the points B, C, D, or E.
     Consider the four points A, B, C, and D. By the supposition and that B, C, and D are noncollinear, A is collinear with two of the points B, C, or D. Without loss of generality assume A, B, and C are collinear. Now consider the four points A, B, D, and E. By the supposition and that B, D, and E are noncollinear, A is collinear with two of the points B, D, or E. If A, B, and D are collinear, then by Axiom P1 and the collinearity of A, B, and C, the points B, C, and D would be collinear, which is a contradiction. Hence, A, B, and D are noncollinear. Similarly, A, B, and E are noncollinear. Thus, A, D, and E are collinear. By Axiom P3, there is a third point F on line BD. Consider the four points A, B, E, and F. By the suppositions at least three must be collinear. If A, B, and E are collinear, then B, C, and E are collinear since A, B, and C are collinear, which is a contradiction. Similarly, each of the sets of points  {A, B, F}, {A, E, F}, and {B, F, E} lead to a contradiction.//