MHB Proving equivalence to Euclid Parallel Postulate

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The discussion centers on proving the equivalence between the Euclid Parallel Postulate and Proclus’s Axiom. The Euclid Parallel Postulate asserts that for any line and a point not on it, there exists exactly one parallel line through that point. In contrast, Proclus’s Axiom states that if two lines are parallel, any line intersecting one must also intersect the other. A user demonstrates that the Parallel Postulate implies Proclus’s Axiom by showing that if a line intersects a parallel line, it cannot be parallel to the other. The conversation invites further exploration of the converse relationship.
pholee95
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I'm having a hard time proving that the Euclid Parallel Postulate is equivalent to this theorem. Can anyone please help?

Euclid Parallel Postulate states: For every line l and point P not on l, there exists exactly one line m so that P is on m and m||l.

the theorem states: (Proclus’s Axiom) If l and l' are parallel lines and t is not equal to l is a line such that t intersects
l then t also intersects l'.
 
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pholee95 said:
I'm having a hard time proving that the Euclid Parallel Postulate is equivalent to this theorem. Can anyone please help?

Euclid Parallel Postulate states: For every line l and point P not on l, there exists exactly one line m so that P is on m and m||l.

the theorem states: (Proclus’s Axiom) If l and l' are parallel lines and t is not equal to l is a line such that t intersects
l then t also intersects l'.
Let's show that the Parallel postulate implies Proclus.

Let $\ell$ and $\ell'$ be parallel lines and $t$ be a line different from $\ell$ which intersects $\ell$. We want to show that $t$ intersects $\ell'$. Say $t$ intersects $\ell$ in a point $p$. If $\ell=\ell'$ then there is nothing to prove. So assume that $\ell\neq \ell'$. So $\ell$ is a line passing through $p$ which is parallel to $\ell'$. By the Parallel Postulate, $\ell$ is the unique such line since $p$ is not on $\ell'$. Thus $t$ cannot be parallel to $\ell'$. Therefore $t$ must intersect $\ell'$ and we are done.

Can you try the converse?
 
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