euclidean-geometry
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.
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).
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]
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]
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]
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.
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]
[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
Your questions seem to have as common theme the underlying question "What is the point of Euclid's postulates?". To answer this, remember that these postulates where introduced in the context of Euclid's book Elements. Therefore, it makes sense to first consider the question "What us the point of Euclid's Elements?".
The point of Euclid's Elements is to collect statements and constructions concerning lines, points and circles in the two-dimensional plane, all of which are known to be absolutely true. To show the reader that these statements are indeed true, Euclid uses a technique which is even today still the basis of all mathematics, namely mathematical proof. This works as follows: you start with a set of statements, all of which you know to be true beyond doubt, and you show that some other statement is a logical consequence of this statement. It then follows that this last statement must also be true beyond doubt. As an admittedly contrived example, if you already know that the statements $x = 3$ and $y =5 $ are both true, then you also know that the statement $x \cdot y = 15$ is true.
So the point of mathematical proof is to expand the collection of statements of which we are absolutely sure. But there is a flaw in this system, namely that we already need to know certain things before we can start proving other things. So before we can use mathematical proof to show which statements are true, we must already have a non-empty set of true statements.
The point of the postulates is exactly to provide this first set of true statements. The point is that we only need to agree on the fact that these five postulates are true, and then the proofs in Euclid's Elements guarantee the truth of all other statements in the book.
So if the postulates feel like they are completely self-evident and almost too obvious to be worth writing down explicitly, then the postulates do exactly what they are supposed to do.
Best Answer
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.