Proving equivalence to Euclid Parallel Postulate

[pholee95]

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'.

 

[caffeinemachine]

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?