Prove without Parallel Postulate Let $x$ and $y$ be parallel lines where $x\neq y$.   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.
 A: 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.
A: 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.
