Prove that Proclus' axiom is equivalent to Playfair's axiom

Question

I'm attempting to prove that Proclus' axiom:

"If a straight line intersects one of two parallel lines, it will intersect the other also."

is equivalent to Playfair's axiom:

"In a plane, given a line and a point not on it, at most one line parallel to the given line can be drawn through the point."

However, before coming to this problem, we've proved Euclid's first $28$ postulates where lines intersected by "magic." Does anyone have an idea/solution to then do this:

You should show that your axiom is equivalent to Playfair's Axiom (so if one holds, so does the other, and vice-versa).

Answer 1

I'll show directly using only the stated axioms and the definition of parallel lines that these are equivalent.

Playfair$\implies$Proclus: Suppose that the Playfair Axiom holds. Let $\ell_1\parallel\ell_2$ and suppose $m$ is a new line intersecting $\ell_1$ in a point $p$. As $\ell_2$ is parallel to $\ell_1$ and $m$ is a line through $p$, we have by the Playfair Axiom that either $m=\ell_1$ or $m\not\parallel\ell_2$. But we assumed that $m\neq\ell_1$, and so we must have $m\not\parallel\ell_2$. By definition of parallel lines $m$ must intersect $\ell_2$. Hence the Proclus Axiom holds.

Proclus$\implies$Playfair: Suppose that the Proclus Axiom holds. Let $m$ be a line with point $p$ not on $m$ and assume to the contrary that there are distinct lines $\ell_1$ and $\ell_2$ through $p$ and parallel to $m$. Now $p$ is on $\ell_1$ and $\ell_2$ with $\ell_2\parallel m$ implies by the Proclus Axiom that $\ell_1$ intersects $m$ in a point $q$. But this contradicts the assumption that $\ell_1\parallel m$. Thus we must have that there is at most one line parallel to $m$ and through $p$. Hence the Playfair Axiom holds.

Answer 2

The axiom can be proven indirectly as well.

Let l and m be two parallel lines, and let t be a transversal that intersects line l. Prove that l also intersects m.

Assume line t does not intersect m. Then m and t are parallel. And lines l and m are parallel. Therefore line t is parallel to line l, as well. But lines l and t were given to not intersect. This is a contradiction. Therefore, t intersects both lines l and m.