Parallel postulate: If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.
Playfair's axiom: Given a line and a point not on it, at most one parallel to the given line can be drawn through the point.
Prove the Parallel postulate from Playfair's axiom.
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$\begingroup$To show Playfair implies Parallel is the same as to show (not Parallel) implies (not Playfair). So assume (not Parallel), i.e. there is a line $t$ (extension of the segment you refer to) intersecting lines $l,l'$ and such that the sum of the interior angles on one of the sides of $t$ is less than $180^{\circ}$ (two right angles), yet $l,l'$ do not meet on that side of $t$. They cannot meet on the other side either, since on that other side the sum of interior angles exceeds $180^{\circ}$, and even in neutral geometry (i.e. with no parallel postulate) the sum of two angles in a triangle cannot exceed $180^{\circ}$. Therefore $l,l'$ are parallel. We may also construct another line $l''$ through the point $P$ on $l'$ where the transversal $t$ meets it, such that for this line the sum of its internal angles on the same side of $t$ is equal to $180^{\circ}$, and this line $l''$ is parallel to $l$ by neutral geometry only. So we now have two distinct parallels to $l$ through $P$ and have arrived at the negation of the Playfair axiom.
$\endgroup$ 7 $\begingroup$History is rich of incorrect proofs about the parallel postulate. We have developed very rigorous proofs checked using the Coq proof assistant.
You can find a computer checked proof here: and the English version of the proofs in the following paper:
You can also find the proof in Greenberg's book "Euclidean and non-euclidean geometries".
$\endgroup$ $\begingroup$I think that the proof can be given directly, which gives it more clarity, using Propositions I.13, I.17, I.28 and I.31 (which do not depend on the Fifth Postulate), following the demonstration in the book Introduction to non-Euclidean geometry by Harold E. Wolfe (Chapter 2).
First, note that the line parallel to a line given by a point not on the line constructed in Proposition I.31 ("To draw a straight line through a given point parallel to a given straight line") produces internal angles of the same side. whose sum is equal to two right angles, this is deduced by virtue of Proposition I.13 ("If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles").
Now, to prove that Playfair Axiom implies the Fifth Postulate it is as follows:
Given lines $AB$ and $CD$ cut by the transversal $ST$ in such a way that the sum of angles $BST$ and $DTS$ is less than two right angles. Construct through $S$ the line $QSR$, making the sum of angles $RST$ and $DTS$ equal to two right angles. This line is parallel to $CD$ by Proposition I.28 ("If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another"). Since lines $QSR$ and $ASB$ are different lines and, by Playfair’s Axiom, only one line can be drawn through $S$ parallel to $CD$, we conclude that $AB$ meets $CD$. These lines meet in the direction of $B$ and $D$, for, if they met in the opposite direction, a triangle would be formed with the sum of two angles greater than two right angles, contrary to Proposition I.17 ("In any triangle the sum of any two angles is less than two right angles").
