The claim made
... that the exterior of a circle is a segment connected set, which means that if $B$ and $C$ are two points outside a circle $\Gamma$, then there exists a third point $D$ such that $\overline{BD}$ and $\overline{DC}$ are entirely outside $\Gamma$.
is not true in neutral plane geometry (a.k.a. the Hilbert plane). More precisely, it is not provable since it is true in Euclidean geometry and false in hyperbolic geometry, both of which are models of neutral geometry.
However, using the definition of a segment-connected set given in Hartshorne's Geometry page 80:
A subset $W$ of the plane is
segment-connected if given any two
points $A, B \in W$, there is a finite
sequence of points $A = A_1, A_2, \dots , A_n = B$
such that for each $i = 1, 2,..., n-1$,
the segment $\overline {A_iA_{i+1}}$ is entirely contained
within $W$,
it can be shown that, in neutral geometry, the exterior of a circle is segment-connected.
Proof that the claim is true in Euclidean geometry
Let circle $\Gamma$ have radius $r$ and center $O$. If the line $\ell$ connecting $B$ and $C$ does not pass through $\Gamma$, we are done. Otherwise, let us construct a line $b$ passing through $B$ perpendicular to $\overleftrightarrow{OB}$ and a line $c$ passing through $C$ perpendicular to $\overleftrightarrow{OC}$ If line $\ell$ does not pass through $O$, $B$ and $C$ are on the same side of a line $m$ parallel to $\ell$ and passing through $O$, and the measure of angle $\angle BOC$ is less than $180$. If we consider $\ell$ a transversal cutting lines $b$ and $c$ and consecutive interior angles $\beta$ and $\gamma$ sum to less than $180$. Thus by Euclid's fifth postulate lines $b$ and $c$ intersect at some point $D$ and because all points on $b$ are at least $OB$ from $O$ and all points on $c$ are at least $OC$ from $O$ the segments $\overline{BD}$ and $\overline{DC}$ are entirely outside $\Gamma$.
If $\ell$ does pass through $O$, then without loss of generality let us assume $OB \leq OC$. Set $\epsilon = \frac{OB-r}{2}$ and construct a line $n$ perpendicular to $\ell$ at $O$. Find a point $O'$ on $n$ a distance $\epsilon$ from $O$. Let us now consider a second circle $\Gamma'$ with center $O'$ and radius $r+\epsilon$. The circle $\Gamma'$ fully contains $\Gamma$ since, if point $X$ is in or on $\Gamma$, $OX \leq r$ and, by the triangle inequality, $O'X \leq OX + \epsilon \leq r+ \epsilon$. So $X$ is in or on $\Gamma'$. Point $B$ is not in $\Gamma'$ since $m$ is perpendicular to $\ell$ at $O$ and $O'B > OB = r + 2\epsilon > r+ \epsilon$. We get a similar result for point $C$. Hence, we can apply the first part of this proof to $B$, $C$ and the new circle $\Gamma'$ to address this case.
Proof that the claim is false in hyperbolic geometry
Consider any circle $\Gamma$ with center $O$ and radius $r$ and with diameter $\overline{QQ'}$. Let $\ell$ be the line perpendicular to line $\overleftrightarrow{QQ'}$ at $O$, and let $P$ and $P'$ be the points of $\ell$ which intersect the circle. Let $m$ be the line passing through $P$ that is asymptotic to line $\overleftrightarrow{QQ'}$ on the $Q$ side of $O$. Drop a perpendicular from $O$ to the line $m$, and let $F$ be the foot of that perpendicular. The point $F$ will be the closest approach that $m$ comes to $O$ and, since $m$ is not perpendicular to $\ell$ at $P$, $OF < r$. Let $R$ be a point of $\overrightarrow{OF}$ between $F$ and the circle.
Let $n$ be the line through $R$ that is asymptotic to $m$ in the other direction. since $m$ and $n$ are asymptotic, any line passing through one, in the direction of the asymptote, passes through the other. So the line $\ell$ passes through both. Consider any one of the points of $\ell$ between $m$ and $n$ — call it $B$. Any line through $B$ that avoids the circle must pass through $n$ and must also pass through $m$ but on the $Q'$ side of $\ell$. It consequently cannot cross $m$ on the $Q$ side and so cannot cross the ray $\overrightarrow{OQ}$. We can make the same argument on the other side of $\ell$ to find points, $B$, near $P$ just outside the circle such that if a line passes through $B$ but not the circle, it cannot intersect ray $\overrightarrow{OQ'}$. There is, apparently, no line through these points that can avoid the circle and go across line $\overleftrightarrow{QQ'}$. Of course, we can make a similar argument about points, $C$, near $P'$ just outside the circle. However, if we are to find a $D$ for a pair of these $B$'s and $C$'s such that $\overline{BD}$ and $\overline{DC}$ are entirely outside $\Gamma$, one or the other of $\overline{BD}$ or $\overline{DC}$ must cross $\overleftrightarrow{QQ'}$.
We have shown that in hyperbolic geometry there are points in the exterior of any circle that require more than two segments to connect them and stay in the exterior. Fooling around with the Poincaré disk, reveals that the larger the circle, the more segments may be needed.
Proof that, in neutral geometry, the exterior of a circle is segment-connected
Consider any circle $\Gamma$ with center $O$ and radius $r$ and any points $B$ and $C$ in the circle's exterior. Construct a perpendicular line $m_1$ to line $\overleftrightarrow{OB}$ at point $B$. Note that all of line $m_1$ lies outside the circle. Find a point $A_1$ on line $m_1$ to one side of $B$ at a distance of $2r$ from $B$ and let angle $\theta$ be the measure of angle $\angle BOA_1$. If $C$ falls on or in the interior of angle $\angle BOA_1$, then the ray $\overrightarrow{OC}$ will intersect $m_1$ at a point $E$. Connect $B$ to $E$ with a segment that lies on $m_1$ and so lies entirely outside $\Gamma$, and then finish by connecting $E$ to $C$ with a segment which will also lie outside since $E$ ad $C$ both lie outside $\Gamma$. If $C$ does not fall in the interior of angle $\angle BOA_1$, connect B to $A_1$, and then find the next point of our segment connection on ray $\overrightarrow{OA_1}$ at distance $OB$ from $O$ and call the point $B_1$.
Now repeat the process with $B_1$. Construct a perpendicular line $m_2$ to line $\overleftrightarrow{OB_1}$ at point $B_1$, and find a point $A_2$ on line $m_2$ to the side of $B_1$ opposite $B$ at a distance of $2r$ from $B_1$. By SAS, the triangle $\triangle OBA_1$ is congruent to $\triangle OB_1A_2$ and so the measure of the angle $\angle B_1OA_2$ is also $\theta$. Again, if $C$ falls on or in the interior of angle $\angle B_1OA_2$, then we can find, as before, a point $E$ on $m_2$ and connect $B_1$ to $E$ to $C$ with segments. If not, connect to $A_2$ and find the next point of our segment connection on ray $\overrightarrow{OA_2}$ at distance $OB$ from $O$ and call the point $B_2$.
We can repeat the process until, by Archimedian property of angle measures, $C$ lies within $\angle B_iOA_{i+1}$ by which time we will have constructed a segment connection between $B$ and $C$ entirely in the exterior of $\Gamma$.
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
Best Answer
Both (i) and (ii) are strictly weaker than the parallel postulate. Probably the easiest way to see this is using coordinates. Given any Pythagorean ordered field $F$ and a convex subring $F_0\subseteq F$, the full subplane $F_0^2$ of the coordinate plane $F^2$ is a Hilbert plane which will always satisfy (i) and (ii), and will not satisfy the parallel postulate unless $F_0=F$. (If $F$ is non-Archimedean, you can always find such an $F_0$ which is not all of $F$, for instance by taking $F_0$ to be the set of all "finite" elements of $F$, i.e. elements bounded by integers.) Conversely, every Hilbert plane satisfying the parallel postulate is isomorphic to the coordinate plane over some Pythagorean ordered field, and so will satisfy (i) and (ii).
Here is a proof that $F_0^2$ satisfies (i) and (ii). For (i), suppose we have a non-vertical line with equation $y=ax+b$. If the slope $a$ is in $F_0$, then $ax\in F_0$ for all $x\in F_0$, and thus we must have $b\in F_0$ for the line to have any points in $F_0^2$ at all, in which case the line intersects every vertical line. On the other hand, if the slope $a$ is not in $F_0$ (or if our line is vertical), then $a$ must be infinitely large (since $F_0$ is a convex subring of $F$, it contains all elements which are bounded by an integer), so $a^{-1}$ is infinitesimal and is in $F$. Rewriting the equation as $x=a^{-1}(y-b)$, the argument from the first case now shows that our line intersects every horizontal line.
Thus, every line in $F_0^2$ either intersects every vertical line or intersects every horizontal line. Given three lines, we may thus assume without loss of generality that two of them intersect every vertical line. Now take any point on the third line, and the vertical line on that point will be a transversal to all three lines. (Unless the third line is itself vertical, in which case you can take its intersection with the first line and draw the line from it to any other point on the second line, choosing the latter point so that the line is not vertical.)
For (ii), suppose we have four (or indeed any finite collection of) points in $F_0^2$. Choose $a<b$ in $F_0$ such that both coordinates of all four points are strictly between $a$ and $b$. Then the right triangle with vertices $(a,a)$, $(a,2b-a)$, and $(2b-a,a)$ contains all four points. (Note that this argument actually requires only that $F_0$ is a subgroup of $F$, not a subring as was needed for (i).)