In Fanos geometry Prove that each point is on exactly 3 line

Solution

1. Method I:

Pick a line l (exists by Axiom 1). Choose any point P not on l (exists by Axiom 3). Since l has 3 points (Axiom 2), joining P to each of them gives three distinct lines through P (by Axiom 4 there is a line through P and each of these points. If one of these lines contained two points of l, it would have to be l by Theorem, but this contradicts the fact that P is not on l.) If there where another line through P, it would not meet l, contradicting Axiom 5, so there are exactly 3 lines through P. This argument takes care of all points not on l, to deal with a point on l, say Q, choose a line through P which does not contain Q and repeat the argument using Q and this line.

2. Method II:

Let P be any point. There are 6 other points in the geometry besides P (Using theorem). P is joined to each by a line (Axiom 4). Since each line contains 3 points (Axiom 2), each of these lines contains two points besides P. Thus, there ar e at least 3 lines through P. Any other line through P would have to contain two points, one on each of two different lines that we have already constructed through P (otherwise Axiom 4 is violated), but this violates Axiom 4. So, there are exactly three lines through P.