# Prove without Parallel Postulate

Let x and y be parallel lines where x is not equal to y.
How do I prove that y is in one of the 1/2 planes , let's call it H of x?
How to prove that one of 1/2 planes of y is contained in H.

Any suggestions or comments are welcome.

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what's m? What's a half plane of a line? –  Robert Israel May 13 '12 at 21:04
Half-plane of a line is probably one of the half-planes of which the line is the boundary, I suspect. –  Cameron Buie May 13 '12 at 21:38
I want to say that it involves Postulate 9: given a line & a plane containing it, the points of the plane that do not lie on the line form 2 sets such that 1. each of the sets is convex and 2. if P is in 1 set and Q is in the other, the segment PQ intersects the line. –  user31284 May 13 '12 at 22:32
That looks to be relevant and useful, but what is m? –  Cameron Buie May 13 '12 at 23:08
I apologize Cameron. I meant the line is x, not m. I confused it with another problem that I was working on earlier regarding about parallel lines where we used m and n. –  user31284 May 14 '12 at 3:20

Start off with $P_1,P_2$ as the half-planes of $x$, $H_1,H_2$ the half-planes of $y$. You want to show that $y$ lies in one of $P_1,P_2$, and that one of $H_1,H_2$ is contained in whichever of $P_1,P_2$ that $y$ lies in.

Are you familiar with proofs by contrapositive or contradiction? I'm going to recommend a contradiction approach for the first part, here.

For the first part, let's suppose that $y$ doesn't lie (fully) in either of $P_1,P_2$, so since $y$ and $x$ don't coincide, then a part of $y$ lies in each of $P_1,P_2$. In particular, there is some point $A$ on $y$ in $P_1$, and some point $B$ on $y$ in $P_2$, yes? What does Postulate 9 say about the segment $AB$ and the line $x$, then? What does that mean about the lines $y$ and $x$, since $AB$ is a segment of $y$, and since $y$ and $x$ aren't the same line? We should have the desired contradiction, here.

For the second, go ahead and assume that $y$ lies in $P_1$ (we've shown it lies in one of $P_1,P_2$, and we can always reindex, if need be), and that $H_2$ isn't contained in $P_1$. All you've got to show is that $H_1$ is contained in $P_1$. I'd start by showing that $x$ must lie in $H_2$--by reasoning as in the first part, $x$ must lie in one of $H_1,H_2$, and we can use our assumption that $H_2$ isn't contained in $P_1$ to show that $x$ must lie in $H_2$. After that, recall that all these half-planes are convex, as is any line in the plane, and note that the overlap of two convex sets is again convex. It will be sufficient (since $x$ lies in $H_2$ and $y$ lies in $P_1$) to show that $H_1$ and $P_2$ have no overlap. Hopefully, that's enough to get you going.

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Thanks for your help, Cameron. –  user31284 May 16 '12 at 3:25

If the half-planes, H1 and H2, are determined by line y. Let point A be in H1 and point B lies in H2. Select points C and D to lie on line y. Should I prove that line y intersects line segment AB to show that A lies in H1 and B lies in H2? And then how do I show that point B is located in a half plane contained by the one that has point A is on the same half plane. Any suggestions or comments are welcome.

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I will give you some guidance in an answer, shortly –  Cameron Buie May 15 '12 at 20:34
Apologies for the delay. I had to go to class. –  Cameron Buie May 15 '12 at 22:56

Let the half-planes of x be called H1 and H2. Let A, B be points not on x. We define an equivalence relation on the complement of x in the plane: A~B if and only if the segment AB crosses x. Clearly this partitions the complement of x into two equivalence classes, namely H1 and H2.

If y is a line parallel to x and A, B are any two points of y, then the segment AB cannot cross x. So A and B are in the same equivalence class, i.e. the same half-plane. Since every point of y is in the same half-plane, we are done.

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