The article in http://www.cut-the-knot.org/fta/Eat/EAT.shtml is dealing with Proposition 16 of the "Elements" of Euclid, which states that:
"In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles."

The importance of this proposition comes clearer once it is used in the proof of Prop. 17:
"In any triangle the sum of any two angles is less than two right angles" (meaning less than 180 degrees).
and in the proof of Prop. 18:
"In any triangle the angle opposite the greater side is greater."
which is then used in the proof of Legendre's Lemma #2:
"In any triangle, the sum of the three angles is less than or equal to two right angles."
(see here: http://www.cut-the-knot.org/triangle/pythpar/PTimpliesPP.shtml#Lemm2).

As we all know, all these (P16,P17,P18 and L2) are all incorrect on the surface of a sphere, where the sum of the three angles of a triangle can grow up to 6 right angles (540 degrees).

In the article above mensioned, the auther (Scott E. Brodie) attacks Euclid's proof of Prop. 16 on this basis, and suggests an alternative proof, adopted by Hilbert (Theorem I.22).

Now, I am not an expert in geometry, hence I beg your pardon if the following is a complete nonsense (or even nonsence). Still, I would like to chalange Brodie's attack and save Euclid's proof as follows.

Euclid's Postulate 1 is this:
"To draw a straight line from any point to any point."

At least two questions can be asked about it's interpretation:
I: Does the postulate speak about straight line segements, or about complete straight lines?
II: Does the postulate only asset that there exists at least such a line, or does it also assert that this line is unique?

Based on the possible answers to these two questions we can more clearly formalize four different postulates:
1a: For any two (different) points there is a complete straight line, going through both of them (at least one).
1b: For any two different points there is a UNIQUE complete straight line, going through both of them.
1c: For any two (different) points there is a straight line segment, going through both of them (at least one).
1d: For any two different points there is a UNIQUE straight line segment, going through both of them.

Surely, 1a implies 1c, and using Euclid's Postulate 2 ("To produce a finite straight line continuously in a straight line."), 1c implies 1a too. Hence, we can consider them equivalent (under Post.2).

Thus we are left with 3 alternatives: 1a, 1b and 1d. Amongst them, one can see that 1b implies 1a, but not vice-versa, and 1d implies 1b (again with the aid of Post.2), but not vice-versa. Hence: 1d > 1b > 1a.

On the "Euclidian plain" all 3 are valid, however, as we know, this is not the case on the surface of a sphere, where, "complete straight lines" are defined to be the big circles that cut the sphere into 2 equal parts (like the meridians/longitudes and the equator, but not the other latitudes). 1d fails, since any two points connected by a segment are also connected by the complement segment (the one which completes the first to a complete line/ring). Furthermore, any two antipodes (e.g. the south and north poles) have (infinitely) many lines in comon (all the meridians in the case of the poles), thus 1b fails too, leaving us only with 1a, as the only acceptable version of the axiom for "Spheric geometry".

However, we can define a Point (note the cap-P) to be the pair of two antipodes. Then, any two "Straight lines" (big circles) share only one Point (the two antipodes) in common. It is easy to see that the statement: "Any two different complete straight lines share only one Point in common" is actually equivalent to 1b (assume one and not the other and you get a contradiction). Hence, under the new definition of Point, 1b holds. This geometry is called "Elliptic geometry". Note though, that 1d does not hold even for Elliptic geometry. Any two different Points are still connected through (exaclly) two complementing Straight segments.

Order (or Betweeness) is a relation amongst points. We say B(x,y,z) if and only if all the following conditions are true:
1. All three points x, y, and z share the same complete straight line;
2. Points x and y share a line segment which doesn't go through z;
3. Points y and z share a line segment which doesn't go through x;
4. Points x and z DO NOT share a line segment which doesn't go through y.
In other words, any straight line segment from x to z goes through y (in the way).

This order relation is defined by the axiom: B(x,y,x) --> x = y. From this we can prove that for different x, y, and z B(x,y,z) --> not B(y,z,x) (or else, B(x,y,z) and B(y,z,x) would imply B(x,y,x) for x != y).

It is easy to realize that in the Eculidian space all points on any straight line are ordered, but that on the surface of a sphere, (assuming either the Spheric geometry or the Elliptic geometry) points (or Points) are never ordered.

This should not come is much surprise, as one can actually prove that 1b Order is equivalent to 1d.

Hence we can characterize the Geometries so far as:

Spheric (Holding only 1a)
Elliptic (Holding 1a and 1b)
Ordered (Holding 1a, 1b and 1d).

Ordered geometries include Euclidian geometry, but also others which do not assume the Parallel postulate (like Hyperbolic geometry).

Now we can come back to Brodie's attack on Euclid's proof of Prop. 16 and his suggestion to use Hilbert's alternative proof. But Hilbert's proof uses 1d, and if we are allowed to assume 1d, Brodie's counter-example to Euclid's proof fails. Brodie's point F fall's not "inside" the angle ACD (as Euclid expects), but on the segment CD. Hence we get two alternative straight routes from B to F, namely: BEF and BCF. But this is possible only for non-Ordered geometries, like the Spheric and Elliptic. If we assume Order, hence 1d, Brodie's counter-example is impossible.

Still one can ask: what if we change Brodie's example, so point F lies even further "below" line CD, completely outside angle ACD? Then supposedly we do have a counter-example to Euclid's proof. This is true, but … (IMHO) this flaw exists in Hilbert's proof too, unless …

It is well known that Euclid's set of postulates was incomplete, and one of the missing postulates is Pasch's axiom. This axiom has quite a few formulations, but it basically asserts that:
"A complete straight line (which is not the infinite extension of any of the triangle sides and) which intersects a side (or two sides at a vertex) of a triangle must then intersect another side as well."
In other words, if the line "enters" the triangle, then it must "exit" as well. This axiom asserts the planarity quallity of all "planar geometries". If the two lines / and \ cross like X, then they must intersect (the cross place is a Point common to both lines). [This axiom doesn't hold if we let a line to "jump over" the cross, or if / and \ are the shadows of lines in 3-dimensional space which do not intersect).

However, if we assume Pasch's axiom, then even the augmented counter-example fails. Line EF in the augmented counter-example cannot exit the angle ACD and point F to fall "below" line B-C-D, without first line EF intersecting the complete stright line going through B-C-D. Call the point of intersection G, and again you have two straight routes: BCG and BEG, contradicting the Order and 1d.

Now, Hilbert's proof need's Pasch's axiom too, or else point E in the second part of the proof might not exist at all.

So, we can conclude that is order to prove Prop.16 (and that there is at least one parallel line to a given line through an "external" point, and that the sum of angles of a triangle is no nore that two right angles = 180 degrees) one needs both Pasch's axiom and 1d, while Pasch and 1b (without Order) are insufficient (as is the case of the Spheric and Elliptic Geometry). Thus, there is no real advantage in Hilbert's proof over Euclid's.

What do you think?