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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).

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  • $\begingroup$ I may be wrong but I thought Playfair's axiom was that if point p is not on line L, there is exactly one line thru p that doesn't intersect L. What def'n of parallel are you using? $\endgroup$ – DanielWainfleet Feb 16 '16 at 0:48
  • $\begingroup$ @user254665 I don't see what difference you are trying to point out between your definition and the one I provided (from Wikipedia) $\endgroup$ – Kevin Feb 16 '16 at 0:52
  • $\begingroup$ In elliptic geometry there are no parallel lines and your version of Playfair holds but mine doesn't. And does parallel mean non-intersecting,or non-intersecting and at a constant distance apart? In hyperbolic geometry there are many of the former but none of the latter. $\endgroup$ – DanielWainfleet Feb 16 '16 at 0:57
  • $\begingroup$ I don't see how mine holds in elliptic geometry if there are no parallel lines, but my axiom claims that there is a parallel line. Anyways, assume Euclidean Geometry, and utilize the first 28 postulates to show the two axioms are equivalent. $\endgroup$ – Kevin Feb 16 '16 at 1:06
  • $\begingroup$ @user254665 It is a theorem of neutral geometry that at least one line may be drawn through a given point not on a given line and parallel to the given line (this was proved by Euclid without reference to the parallel postulate). The Playfair Axiom asserts that such a line is unique. Elliptic geometry is a whole other animal, and I think that our friend here assumes only axioms of neutral geometry to prove this equivalence. $\endgroup$ – User12345 Feb 16 '16 at 1:11
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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.

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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.

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