A Proof with Intersecting Lines My geometry teacher has given me a question to try to solve which is:
Prove that there exists lines a and b, such that a is not equal to b and a intersects b.
I am not sure how to prove this or which proof strategy I need to use.  Any hints would be very nice.
Thank you in advance.
 A: Hint: You need only three non-colinear points. And you only need to prove the existence of two intersecting lines.
That means: Choose any two points, draw the line through the line segment determined by those two points, and then draw a third point not on the first line. The only line through that third point that won't intersect the first line would be a line parallel to the first line. Any other line through the third point will necessarily intersect the first line. 
For a fully Euclidean proof: See the Parallel Postulate:


"The parallel postulate: If two lines intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough."

A: Hint:
Use the Incidence Postulates for three points to have a three point model in this extended consistent geometry. So you have:

Points: $A,~B,~C$.
Lines: $\{A,B\},~\{A,C\},~\{B,C\}$

Can you find your lines from above?
This fig is from Google:

See this file also.
A: When a proof ask you to show that "there exists,"  your goal is simply to find an example.  Perhaps your instructor just wants to make sure that you know what intersecting means.  In any case, just choose your favorite non-equal intersecting lines, and present them with a flourish!  Of course, we have to be talking about lines in two or three or higher dimensional spaces if it is going to be possible to find two non-equal intersecting lines.
