Which statements are equivalent to the parallel postulate?

Solution 1:

Here the list organised by main subject.

Add more if you know some, but add reference to where it comes from.

If a proposition falls under more than 2 subjects you may add them under both. Like triangle 5 ( Every triangle can be circumscribed ) and circle 1 ( Given any three points not on a straight line, there exists a circle through them).

Lines:

Euclid: 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. [1]

There is at most one line that can be drawn parallel to another given one through an external point. (Playfair's axiom)[1,6]
There exists a pair of straight lines that are at constant distance from each other.[1]
Two lines that are parallel to the same line are also parallel to each other.[1,6]
If a line intersects one of two parallel lines, both of which are coplanar with the original line, then it also intersects the other. (Proclus' axiom) [1,6]
if two straight lines are parallel, they are figures opposite to (or the reflex of) one another with respect to the middle points of all their transversal arguments.(Veronese) [2]
Two parallel straight lines intercept, on every transversal which passes trough the middle point of a segment included between them, another segment the middle point of which is the middle of the first (Ingami) [2]
Two straight lines that intersect one another cannot be parallel to a third line. (no 7 at [3] )
If two lines are parallel , then alternate internal angles cut by an transversal are congruent (converse alternate internal angle theorem). [4,6]
If t is a transversal to $l$ and $ l \parallel m $ and $ t \bot l $ then $t \bot m $. [4,6]
if $ k \parallel l $ , $ m \bot k $ and $ m \bot l $ then either $ m=n $ or $ m \parallel n.$ [4]
Any two parallel lines have two common perpendicular lines. [5]
Any three distinct lines have a common transversal. [5]
There are not three lines such that any two of them are in the same side of the third. [5]
Two any parallel lines have a common perpendicular. [5]
Given $r,s$ lines, if $r$ is parallel to $s$, then $r$ is equidistant from $s$.[5]
Given a line $r$, the set of the points that are on the same side of $r$ and that are equidistant from $r$, is a line. [5]
Given lines $r,s,u,v$, if $r$ is parallel to $s$, $u$ is perpendicular to $r$ and $v$ is perpendicular to $s$, then $u$ and $v$ are parallel. [5,6]
Given lines $r,s,u,v$, if $r \perp s$, $s \perp u$ and $u \perp v$, then $r$ cuts $s$ (Bachmann Lottschnitt axiom). [5,6]
If $\overleftrightarrow{AB} \parallel \overleftrightarrow{CD}$ and $\overleftrightarrow{BC}$ is transversal to both of them such that $A$ and $D$ are in the same side of $\overleftrightarrow{BC}$, then $m(\measuredangle ABC) + m(\measuredangle DCB) = 180°$. [5]
For any point P, line l, with P not incident with l, and any line g, there exists a point G on g for which the distance to P exceeds the distance to l [8]

Triangles:

The sum of the angles in every triangle is 180° (triangle postulate).[1,6]
There exists a triangle whose angles add up to 180°.[1,6]
The sum of the angles is the same for every triangle.[1]
There exists a pair of similar, but not congruent, triangles.[1,6]
Every triangle can be circumscribed.[1,6]
In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides (Pythagoras' Theorem).[1]
There is no upper limit to the area of a triangle. (Wallis axiom)[1]
Given a triangle $\Delta ABC$, if $(AC)^2 = (AB)^2 + (BC)^2$, then $\angle B$ is a right angle. (converse of Pythagorean Theorem) [5]
Given a triangle $\Delta ABC$, exists $\Delta DEF$ such that $A \in \overline{DE}$, $B \in \overline{EF}$ and $C \in \overline{FD}$. [5]
Given a triangle $\Delta ABC$, if $D$ and $E$ are respectively the middle points of $\overline{AB}$ and $\overline{AC}$, then $DE = \frac{1}{2}BC$. [5]
(Thales) Given a triangle $\Delta ABC$, with $B$ in the circle of diameter $\overline{AC}$, then $\angle ABC$ is a right angle. [5,6]
The perpendicular bisectors of the sides of a triangle are concurrent lines. [5,6]

Rectangles:

There exists a quadrilateral such that the sum of its angles is 360°. (answer Ivo Terek below)
If three angles of a quadrilateral are right angles, then the fourth angle is also a right angle.[1,6]
There exists a quadrilateral in which all angles are right angles.[1,6]
The summit angles of the Saccheri quadrilateral are 90°. [1,6]
If in a quadrilateral 3 angles are right angles, the fourth is a right angle also.[2,6]

Circles:

Given any three points not on a straight line, there exists a circle trough them. (Legendre, Bolay)[2,6]
A curve of constant non-zero curvature is a circle.
A curve of constant non-zero curvature has finite extent.
There exist circles of arbitrarily low curvature.
The area of a circle grows at most polynomially in its radius.

Other:

Through any point within an angle less than 60° a straight line can always be drawn which meet both sides of the angle. (Legendre)[2]
Given an angle $\angle ABC$ and $D$ in its interior, every line that passes throuh $D$ cuts $\overrightarrow{BA}$ or $\overrightarrow{BC}$. [5,6]
If $A,B$ and $C$ are points of a circle with center $D$ such that $B$ and $D$ are in the same side of $\overleftrightarrow{AC}$, then $m(\measuredangle ABC) = \frac{1}{2}m(\measuredangle ADC)$. [5]
Given a acute angle $\angle ABC$ and $D \in \overrightarrow{BA}$, $D \neq B$, if $t$ contains $D$ and is perpendicular to $\overleftrightarrow{AB}$, then $t$ cuts $\overrightarrow{BC}$. [5]

References:

[1]: wikipedia http://en.wikipedia.org/wiki/Parallel_postulate

[2]: Heath's "Euclid, The Thirteen Books of The Elements" Dover edition

[3]: cut the knot http://www.cut-the-knot.org/triangle/pythpar/Fifth.shtml

[4]: Greenberg's "Euclidean and Non-Euclidean geometries" 3rd edition 1994

[5]: Professor Sergio Alves' notes of Non-Euclidean Geometry, from University of São Paulo (the original notes (in portuguese) in three images: here, here and here)

[6]: The computer checked proofs of the equivalence between 34 statements: http://geocoq.github.io/GeoCoq/html/GeoCoq.Meta_theory.Parallel_postulates.Euclid_def.html and the paper : https://hal.inria.fr/hal-01178236v2

[7]: Martin, The foundations of geometry and the non euclidean plane.

[8]: Pambuccian, Another equivalent of the Lotschnittaxiom, V. Beitr Algebra Geom (2017) 58: 167. doi:10.1007/s13366-016-0307-5

Solution 2:

A few more:

The sum of the angles in every quadrilateral is $360^\circ $.
Exists a quadrilateral such that the sum of its angles is $360^\circ $.
If two parallel lines are cut by a transversal line, then the alternate angles are congruent.
Given lines $r,s,t$, if $r$ is parallel to $s$ and $t$ cuts $r$, then $t$ cuts $s$.
Given lines $r,s,t$, if $r$ is parallel to $s$ and $s$ is perpendicular to $t$, then $t$ is perpendicular to $s$.
Given lines $r,s,u,v$, if $r$ is parallel to $s$, $u$ is perpendicular to $r$ and $v$ is perpendicular to $s$, then $u$ and $v$ are parallel.

Your $13$ is directly equivalent to saying that retangles exist. Remember that a Saccheri quadrilateral ${\bf S}ABCD$ is such that $\overline{AB} \cong \overline{CD}$ and $m(\measuredangle A) = m (\measuredangle D) = 90^\circ$. It can be proved also that $\overline{AD} \parallel \overline{BC}$ and $\measuredangle B \cong \measuredangle C$, among other stuff. A Lambert quadrilateral is one that have three right angles.

I suggest that you look around for neutral geometry, which consists os results that independ of the fifth postulate. A bit of hyperbolic geometry might be useful too, so you can detect where the uniqueness of the parallel was being used.

About the equivalences that use the notion of angle, in neutral geometry they are expressed using the notion of the defect of a triangle or a convex quadrilateral (I believe "defect" is the term used, since english is not my native language and I haven't seen any material about it in english) The defect of $\Delta ABC$ and $ABCD$ are, by definition: $$\delta (\Delta ABC) = 180 - (m(\measuredangle A) + m (\measuredangle B) + m (\measuredangle C))$$ and $$ \delta ( ABCD) = 180 - (m(\measuredangle A) + m (\measuredangle B) + m (\measuredangle C) + m (\measuredangle D))$$

Defect has a few additive properties, and sorta plays the role of "area" in non-Euclidean geometry. So in this fashion, you would have:

2. For every triangle $\Delta$, we have $\delta (\Delta) = 0 $.
3. There exists a triangle $\Delta$ such that $\delta(\Delta) = 0$.
4. Given any two triangles $\Delta_1$ and $\Delta_2$, we have $\delta (\Delta_1) = \delta (\Delta_2)$.

I will come back and add more equivalences here hopefully soon.

EDIT.: More equivalences:

If $\overleftrightarrow{AB} // \overleftrightarrow{CD}$ and $\overleftrightarrow{BC}$ is transversal to both of them such that $A$ and $D$ are in the same side of $\overleftrightarrow{BC}$, then $m(\measuredangle ABC) + m(\measuredangle DCB) = 180º$.
If $\overleftrightarrow{AB} // \overleftrightarrow{CD}$ and $\overleftrightarrow{BC}$ is transversal to both of them such that $A$ and $D$ are in opposite sides of $\overleftrightarrow{BC}$, then $\measuredangle ABC \cong \measuredangle DCB$.
The perpendicular bisectors of the sides of a triangle are concurrent lines.
Given three non-collinear points, exists a circle which contains them.
Exists a unique point equidistant from three non-collinear given points.
Given a acute angle $\angle ABC$ and $D \in \overrightarrow{BA}$, $D \neq B$, if $t$ contains $D$ and is perpendicular to $\overleftrightarrow{AB}$, then $t$ cuts $\overrightarrow{BC}$.
(Legendre) For every acute angle $\angle ABC$, and $D \in \mathrm{int}(\angle ABC)$, exists a line passing through $D$ that cuts both $\overrightarrow{BA}$ and $\overrightarrow{BC}$ in points distinct from $B$.
(Thales) Given a triangle $\Delta ABC$, with $B$ in the circle of diameter $\overline{AC}$, then $\angle ABC$ is a right angle.
If $A,B$ and $C$ are points of a circle with center $D$ such that $B$ and $D$ are in the same side of $\overleftrightarrow{AC}$, then $m(\measuredangle ABC) = \frac{1}{2}m(\measuredangle ADC)$.
If $\angle ABC$ is a right angle, then $B$ is in the circle of diameter $\overline{AC}$.
The perpendicular bisectors of the catheti of a right triangle are concurrent lines.
Given lines $r,s,u,v$, if $r \perp s$, $s \perp u$ and $u \perp v$, then $r$ cuts $s$.
Exists an acute angle $\angle ABC$ such that every line that contains $D \in \overrightarrow{BA}$, $D \neq B$, and it is perpendicular to $\overleftrightarrow{AB}$ also cuts $\overrightarrow{BC}$.
Exists an acute angle such that for every point in its interior, passes a line through it that cuts both sides of the angle in points distinct from its vertex.
Given $r,s$ lines, if $r$ is parallel to $s$, then $r$ is equidistant from $s$.
Given a line $r$, the set of the points that are on the same side of $r$ and that are equidistant from $r$, is a line.
Exists two distinct lines that are equidistant.
(Wallis) Given a triangle $\Delta ABC$ and a line segment $DE$, exists a point $F$, with $D,E$ and $F$ non-collinear such that $\Delta DEF \cong \Delta ABC$.
Exists two triangles that are similar, but not congruent.
There is no absolute measure system.
Any two parallel lines have two common perpendicular lines.
The diagonals of a Saccheri quadrilateral intersect in their midpoints.
Any three distinct lines have a common transversal.
There are not three lines such that any two of them are in the same side of the third.
Given a triangle $\Delta ABC$, if $D$ and $E$ are respectively the middle points of $\overline{AB}$ and $\overline{AC}$, then $DE = \frac{1}{2}BC$.
Two any parallel lines have a common perpendicular.
Given a triangle $\Delta ABC$, exists $\Delta DEF$ such that $A \in \overline{DE}$, $B \in \overline{EF}$ and $C \in \overline{FD}$.
Given an angle $\angle ABC$ and $D$ in its interior, every line that passes throuh $D$ cuts $\overrightarrow{BA}$ or $\overrightarrow{BC}$.
The Pythagorean Theorem.
Given a triangle $\Delta ABC$, if $(AC)² = (AB)² + (BC)²$, then $\angle B$ is a right angle.
Exists a triangle such its area is greater then any value given.

That's all I've got now.

EDIT: The original reference (in portuguese) in three images: here, here and here.