Projective geometry question: 4 points no 3 on a line

[anniecvc]

Homework Statement

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.

Homework 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"

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?

 

[Dick]

Homework Statement

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.

Homework 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"

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?

You should follow your first aha! ABE are on a line, and BCE are on a line. Both lines contain B and E. If B and E are different, then they must be the same line by axiom 1), yes? Wouldn't that mean that A,B and C are on the same line? It does say, no three of which are on a "line."

 

[certainly]

For part b):- show from part a) that the only common point between any 2 of the 4 lines is the common endpoint. Therefore AB and CD have no common point. In fact if you choose any 3 of the four lines you will find that there is one line which does not share any of it's endpoints with one another. you can check this by hand, there are only 4 possibilities.
FOOD FOR THOUGHT:- If this were not so it would violate the fact that not 3 points should be in a straight line in this system. Can you see why?
Cheers.

 

[anniecvc]

You should follow your first aha! ABE are on a line, and BCE are on a line. Both lines contain B and E. If B and E are different, then they must be the same line by axiom 1), yes? Wouldn't that mean that A,B and C are on the same line? It does say, no three of which are on a "line."

Ah, got it. Thank you so much!