2
$\begingroup$

Euclid's Proposition I.32 says that the sum of the internal angles of any triangle is equal to the sum of two right angles.

However, there is a gap in Euclid's proof -- it does not show that the ray $CE$ is between the rays $CA$ and $CD$. Actually, the choice of one of the two available directions for the ray $CE$ (to be parallel to the line $AB$) is not explicitly specified by Euclid.

Here is the illustration for the proof in EUCLID’S ELEMENTS OF GEOMETRY, where I marked equal angles with the same colors:

enter image description here

Is there a published complete synthetic proof of this result based on, say, Hilbert's axioms?

This question is related to this previous answer.

Improve this question
$\endgroup$
10
  • 2
    $\begingroup$ Since you mention the proof using Hilbert's axioms, you might want to take a look at the proof in M. Greenberg's Euclidean and Non-Euclidean Geometries: Development and History. Of course, he pays careful attention to betweenness issues throughout the development. $\endgroup$ Commented Jul 22 at 18:26
  • $\begingroup$ There is a teacher's manual for Greenberg's fourth edition. It includes a list of some 70 hypotheses equivalent to the parallel postulate. $\endgroup$ Commented Jul 22 at 18:34
  • $\begingroup$ @RobertBryant : I had looked at Greenberg's book. The sum-of-angles theorem is Proposition 4.11 in that book. However, the 6-line proof of that proposition there does not mention any betweenness. Its last sentence just says "But the three angles at vertex В have degree measures adding to 180°." $\endgroup$ Commented Jul 22 at 19:01
  • 1
    $\begingroup$ @IosifPinelis: I don't have a copy of the book with me, but I think that I remember that he constructs the angles in question using alternate interior angles, which has the necessary betweenness criteria built in. $\endgroup$ Commented Jul 22 at 19:40
  • $\begingroup$ @IosifPinelis: You have to go back to the sections of the book on betweenness, where Greenberg proves all the little lemmas about half-planes and betweenness for rays to see that his argument absolutely does take care of these issues. When I teach the betweenness lemmas and propositions, the students often have a lot of trouble understanding why he is so careful about betweenness, but by the time we get to Hilbert's Euclidean Parallel Postulate, this has become second nature, and the issues of betweenness for rays that you are worried about are clear. $\endgroup$ Commented Jul 22 at 22:56

2 Answers 2

Reset to default
1
$\begingroup$

Well, one has to say that $E$ and $A$ lie on one side of $(BC)$. Now, since $(AB) \parallel (CE)$, we have that $A$ and $D$ lie on opposite sides of $(CE)$. So $[CE)$ lies between $[CD)$ and $[CA)$.

So, it is not a true gap; it is just style. If Birkhoff's axioms are okay for you, then check 7C in my book.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Can you explain your notations here, such as $(AB)$ and $[CA)$? $\endgroup$ Commented Jul 22 at 18:11
  • $\begingroup$ $(AB)$ is a line and $[AB)$ is a half-line. $\endgroup$ Commented Jul 22 at 18:25
  • $\begingroup$ I think the condition that $A$ and $D$ lie on opposite sides of $(CE)$ does not by itself imply that $[CE)$ lies between $[CD)$ and $[CA)$ -- consider e.g. the case when all pairwise angles between the rays $[CA)$, $[CD)$, and $[CE)$ equal $2\pi/3$. $\endgroup$ Commented Jul 22 at 18:52
  • $\begingroup$ Also, as I requested, do you know a published complete synthetic (rather than metric-based) proof of this result? $\endgroup$ Commented Jul 22 at 18:54
  • 1
    $\begingroup$ Since $A$ and $D$ are on opposite sides from $(CE)$, there is $X \in [AD] \cap (CE)$, and since $E$ and $A$ are on one side of $(BC)$, we get that $X\in [CE)$. + I just checked Hilbert's book; he says it can be proved in the usual way in a footnote in § 14. $\endgroup$ Commented Jul 23 at 4:49
0
$\begingroup$

At this point, the best known to me way to fix the gap in Euclid's proof of Proposition I.32 is by using the exterior angle (EA) theorem presented as Theorem 4.2 in the book by Greenberg; the latter theorem seems to be proved fully and correctly in that book. On p. 165, Greenberg states: "The exterior angle theorem [...] was the 16th proposition in Euclid's Elements. Euclid's proof had a gap due to reasoning from a diagram. He considered [...] He then assumed from the diagram that B' lay in the interior of [a angle]". So, that was another case when Euclid reasoned from a picture and missed a betweenness relation.

Greenberg's EA theorem states: "In any Hilbert plane, an exterior angle of a triangle is greater than either remote interior angle."

So, the fix of Euclid's proof of Proposition I.32 is as follows. Let the ray $CE$ be parallel to the line $AB$ and such that the (blue) angles $BAC$ and $ACE$ in the diagram in the OP be alternate interior angles wrt to the lines $AB$ and $CE$ and the transversal $AC$. Then, by the EA theorem, the angle $ACD$ will be greater than the angle $BAC$, which latter is equal to the angle $ACE$. So, the angle $ACD$ is greater than the angle $ACE$, which shows that the ray $CE$ is between the rays $CA$ and $CD$, as desired.

Remark: As was noted in a comment, 'The sum-of-angles theorem is Proposition 4.11 in [Greenberg's] book. However, the 6-line proof of that proposition there does not mention any betweenness. Its last sentence just says "But the three angles at vertex В have degree measures adding to 180°".'

The proof of Proposition 4.11 Greenberg's book is illustrated there by the following picture:

enter image description here

Missing in that proof is the verification of the facts that (i) the ray $BA$ is between the left dashed ray and the ray $BC$ and (ii) similarly, the ray $BC$ is between the ray $BA$ and the right dashed ray.

Despite the betweenness gap, Euclid's proof of his Proposition I.32 -- requiring the betweenness relation just for one ray $CE$ instead of the two rays $BA$ and $BC$ -- seems more elegant than Greenberg's proof of the sum-of-angles theorem.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.