CHAPTER IV

Proof of Euclid’s Postulate

The state of the question with regard to Euclidean geometry revolves around the proof of Euclid’s postulate, known as the parallel postulate. If the parallel postulate can be firmly established, Euclidean geometry holds; if not, Euclidean geometry will not be free from attack, and non-Euclidean Geometries will continue to flourish, now that they have started. There are two questions to be answered in regard to this matter. Is this postulate itself self-evident, as are the others, against which no valid objection can be urged? and if not, is it capable of being apodictically connected with the other accepted assumptions of Euclid? The decision then between Euclidean and non-Euclidean geometry comes down to the settlement of these two questions.

First, is the so-called parallel postulate an assumption valid without proof? From what we have already seen, we can agree with all the non-Euclidean Geometers, and all those who have attempted to prove it from the days of Euclid to our own, that it is not a proper postulate. In the first place, the proposition itself is too different in simplicity of statement and evidence by intuition from the other postulates to be considered a postulate like them, and is more in the form of a theorem to be proved. In the second place, propositions that are more self-evident than this are actually proved in Euclid’s Elements, and if this one lacks proof, it was because none was found available. As we have shown, there is evidence making it clear that this proposition was a difficulty for Euclid, and for geometers even before his time. Moreover, Euclid himself proves the converse of this postulate in I, 17, where it is shown that in any triangle two angles taken together in any manner are less than two right angles. If the converse requires proof, then the postulate itself is a theorem that requires demonstration. For a postulate should be such that the connection between subject and predicate ought to be clear and self-evident no matter what way they are turned. If there is an essential and evident connection between the formation of triangles and base angles that together are less than two right angles, then both the proposition concerning this connection and its converse will be evident. The fact that Euclid proved the one, and assumed the other, can only point to one thing, his inability to furnish proof for the other. But in the proof of I, 17 Euclid does not give the actual cause of the connection; it is rather a proof that the connection exists. For the reason does not lie in the relation to the exterior angle, but in the necessary convergence to a point that is in the sides of a triangle in the case of two lines forming with a base less than two right angles. Even I, 17 is then not a proper demonstration. It is what Aristotle calls demonstration hoti and not a demonstration dioti which is the proper mathematical demonstration. For there is a great difference between knowledge “that” a thing is, and “why” it is.¹ That Euclid had recourse to this shows that he was hard put to prove his theory. But it shows at the same time, that the theory required proof, and that for the postulate as well as for its converse. The statement of Proclus given as motto to this chapter can then be adopted in its entirety: “This should be stricken out of the postulates entirely, for it is really a theorem.”

The second question, can this postulate be deduced from other valid assumptions, and thus be logically and necessarily connected with the system that goes by Euclid’s name, is the real point at issue. The possibility of doing this is of course denied by all non-Euclidean Geometers, for otherwise their systems would have no logical standing. If this question can be answered in the affirmative, then that other connected question will also be definitely settled, viz., can assumptions contradictory to this postulate be made, and a logical body of geometric doctrine consistent with the other accepted assumptions and deductions of Euclid still be possible, a thing which the new Geometers affirm, and which we shall now proceed to show completely false.

The question as ordinarily placed is, can any assumption be made contradictory to one or other of Euclid’s assumptions and the possibility of constructing a logical and certain doctrine still be retained. But this method of putting the question depends on the further assumption that the postulate of Euclid is not capable of proof, and this we are going to show to be false and without foundation. By answering the direct question above, and proving directly the parallel theory, the whole question will be automatically settled. For if non-Euclidean Geometry is true, the postulate cannot be demonstrated. The converse is likewise true, if the postulate can be strictly demonstrated, the non-Euclidean Geometries have no logical grounds for existence. If then we can prove the Euclidean postulate, Euclidean geometry still stands, and the non-Euclidean Geometries will necessarily fall.

The immediate question for us, therefore, is where does the truth lie, and where do the geometers lie. If we can establish both postulate and definition, as we claim and shall incontestably prove, then we have Euclid’s geometry solidly established. If the parallel definition alone be admitted or what is sometimes called the sixth postulate, then we have Lobatschewsky’s Geometry. If we admit the fifth postulate and refuse the parallel definition as representing reality, we have Riemann’s Geometry.

Let us then without more ado examine the fifth postulate, to see if we can validly establish it, for on this one thing the whole question will be decided. This postulate is the converse of the theorem that the sum of the angles at the base of a triangle is less than two right angles, which itself is but a corollary to the proposition that the sum of all the interior angles of a triangle is equal to two right angles. We have already seen how different mathematicians recognized this necessary connection, and made various efforts to establish this proposition without having recourse, as does Euclid, to the famous disputed postulate to accomplish it. This is what we profess to do.

This proposition, as we shall see, immediately admits of direct demonstration, and not a mere reductio ad absurdum, a method which has weakened the position of nearly all those who attempted its proof, and at the same time misled them. Furthermore, as a matter of method, it should be as capable of direct proof as any other of the fundamental theorems of geometry. A large part of the difficulty in proving the postulate throughout the ages came from the preconceived notion of the difficulty of the proof, which arose only from the originally conceived imperfect order in the theorems of Euclid. As we shall now show, when taken in its proper order, it is just as capable of simple and easy proof as any other proposition in Euclid.

Furthermore, it can be established without assuming anything that is either explicitly or implicitly contained either in the postulate or in the definition of parallels. Our method then does not consist, as in all the other demonstrations, of assuming any other supposedly simpler postulate, but by establishing the postulate as a regular theorem, and using only the other postulates of Euclid. As a matter of fact, we shall prove it without making any further assumption, or using any proposition, other than those already contained in the axioms, postulates (excluding the fifth, or the one we intend to prove), and theorems previous to the first use of this postulate in the first book of Euclid’s Elements. To be strictly accurate, our demonstration will contain nothing not found in the first fifteen propositions of Euclid’s first book. The proof is geometric simplicity itself. There is nothing in it that need cause either hesitation or doubt. It is all a question of proper order in proving, a thing that has hitherto vitiated the treatment of this question.

We shall, therefore, omit the sixteenth proposition of Euclid as being useless. This proposition is, “If one side of a triangle be produced, the exterior angle is greater than either of the interior and opposite angles.” For, since we intend to prove that it is actually equal to the sum of these, which Euclid does later in the thirty-second, but by making use of the parallel postulate, it has no other Purpose than to be a stepping stone on the road to the other. In this sense it is not necessary for our proof, and is better omitted. As we saw before, the sixteenth is but an inchoative and indefinite thirty-second, and would only have been introduced by one who could not prove the complete proposition without the parallel assumption. This defect we now undertake to remedy.

We shall also omit the thirty-third and the thirty-fourth propositions, as being replaced by equivalent ones, and also change their order. Then substituting new theorems for those omitted, changing the proof of some of the others, and further altering somewhat the order of Euclid as far as this question is concerned, we shall proceed with our demonstration. Our first proposition will cover somewhat the same ground as that of the thirty-third and thirty-fourth, and will consist in the construction of a parallelogram.

The parallelogram is not among the definitions of Euclid, and accordingly we shall supply it.

A parallelogram is a four sided figure having its opposite sides and angles equal. We shall also change Euclid’s definition of parallel lines to correspond with our construction.

Parallel lines are lines that lying in the same plane are equidistant at equidistant points. We shall omit all discussion of this definition till we treat the question of parallel lines by itself, when we propose to justify it completely.

Our first proposition then will be:
Given a triangle, to construct a parallelogram.
Let ABC be any triangle.
Bisect the base AC at E, and from B through E draw the line BF, such that BE equals EF.
Join AF and FC.

Then ABCF is the required parallelogram.

Proof. BE = EF,
          AE = EC,
(Both by construction.)

Angle BEA = Angle CEF.
(Both vertical angles.)

Therefore t r i a n g l e ABE triangle CEF.
(Having two sides and included angle of the one equal to two sides and included angle of the other.)

Therefore the side FC = the side AB.
(Corresponding sides of equal triangles.)

Similarly, the triangles BEC and AEF can be proved equal.
Therefore BC = AF.
But angle ABE = angle EFC.

And angle CBE = angle AFE.
(Corresponding angles of equal triangles.)

Therefore angle ABC = angle AFC. (Ax.2).²
Similarly angle BAF = angle BCF.

Therefore ABCF is a parallelogram.
(Having opposite sides and angles equal.) Q. E. D.

This proof is very simple and clear and in the strictest geometrical method. From this the parallel theory is but a simple deduction. But first we shall deduce an evident corollary contained in the above problem. It is this:

Cor. In every parallelogram the alternate-interior angles formed by the diagonals with any pair of opposite sides are equal; e.g., angle BAC equals angle ACF.

So far we have shown that the opposite sides of a parallelogram are equal, and hence that opposite sides are equidistant at the equidistant points that are their extremities. But this is not enough to show that these lines are parallel, that is, that they are everywhere equidistant. This we shall now prove, and by a new proposition.

PROPOSITION II

If a transversal falling on two straight lines makes the alternate-interior angles equal, any equal segments of the two lines can be made the sides of a parallelogram about the transversal as diagonal.

Let AB and CD be the two straight lines, and EF the diagonal intersecting AB and CD at E and F. Also let angle AFE equal angle FED.

To prove that any equal segments of AB and CD can be made the sides of a parallelogram about EF.
Proof. Take any other point H on the line CD.
On AF take FG equal to EH. Join FH and EG.
Then EH equals GF (Const.).
EF is common.
And angle GFE equals angle FEH (Hyp.).
Therefore triangle GFE equals triangle FEH (Prop. IV).
And GE equals FH; angle EGF equals angle FHE; and angle GFH equals angle HEG (Ax. 2).
Therefore, GFHE is a parallelogram. Q. E. D.

We have now in two propositions, each of them clear and apodictic, completely established the basic doctrine of the parallel theory, the reality of parallel lines. They are lines that are equidistant at equidistant points.

They are equidistant not only at the extremities of the sides of the parallelogram, a thing which was proved in the first proposition, but anywhere on either line. For we may take any point G to establish the parallelism, and therefore it is proved for every point. There is only one condition required, and that is that the two lines be such that the angles formed with them by a transversal, and called the alternate-interior angles, be equal. This being given, we may take any other points on either line equidistant from the two points of intersection, and show that the pair of points in one line is equidistant from the corresponding points of the pair on the other line. In other words, the complete figure will form a parallelogram.

This equidistance everywhere between any points of the opposite sides of a parallelogram may be shown in another way, and for this we shall deduce another evident corollary from the first proposition. It is as follows:

Cor. The diagonals of a parallelogram bisect each other; or, the point of intersection of two diagonals of a parallelogram bisects the diagonals.

This has already been shown in the proof of Proposition I. We shall now extend this further to include all transversals, in a new theorem of easy proof, which shall read as follows:

PROPOSITION III

The point of intersection of the diagonals of a parallelogram bisects every straight line passing through it and terminated in opposite sides of the parallelogram.

Let ABCD be the parallelogram, and AC and BD its diagonals intersecting at O; let EF be any other line passing through O and meeting AB and CD in E and F.

To prove OE equals OF.

Proof. OA equals OC (previous corollary). Angle OAE equals angle OCF (Alt. Int. angles). Angle AOE equals angle COF (vert. angles). Therefore triangle AOE equals triangle COF (Prop. 26). And OE equals OF. Q. E. D.

If now we join AF and EC we have two new triangles, OEC and OAF, which are easily shown to be equal. For the two sides are already proved to be equal in the proposition above, and the included angles, being vertical angles, are also equal. Therefore, by Prop. IV the triangles are equal, and hence the sides AF and EC are equal; or any pair of points we wish to choose on one line is shown to be equidistant from an equidistant corresponding pair on the other line, whether the distance be measured parallel to the sides of the original parallelogram or not.

The conclusion again follows with absolute necessity, that the opposite sides of a parallelogram are parallel, since they are everywhere equidistant at equidistant points. It makes no difference where the equidistant points are taken, or in what direction from one another. Likewise in proposition II the points may be chosen anywhere on either line indefinitely produced. They are not restricted to any given parallelogram. There is no doubt or hesitation possible as to the apodictic demonstrating power of the arguments, or of the geometric necessity of the conclusion. The arguments are geometrically rigorous.

With this much proved the rest of the parallel theory follows. It is evident that when lines are parallel the alternate-interior angles are equal; also the proofs hold of the equality of the exterior-interior angles, and that the sum of the interior angles on the same side of the transversal is equal to two right angles as given by Euclid. But these can easily be proved by the following or fourth proposition. The converse is also easily proved in the same manner in the fifth proposition. Having finished with these proofs, we can immediately enunciate the proposition on which depends the parallel postulate, Euclid I, 32, about the interior angles of a triangle, which will then follow in all rigor.

PROPOSITION IV

A straight line falling on parallel straight lines makes the alternate-interior angles equal to one another, the exterior angle equal to the opposite and interior angle, and the interior angles on the same side of the transversal equal to two right angles.

(1)

Let AB and CD be two parallel lines, and let EF be a third straight line cutting them transversely. Take any point G on AB between A and F.
Join EG.
Take point H on the line CD between E and D, such that EH equals GF.
Join H and F.
Then GFEH is a parallelogram having its opposite sides AB and CD parallel.
Proof.
For GF = EH (by construction).
EF is common.
GE = FH (parallel lines are equidistant at equidistant points).
Therefore triangle GFE = triangle FEH.
And angle GFE = angle FEH (Prop. 8).
Similarly it can be proved that angle BFE = angle CEF. For GFE + EFB = FEH + FEC (both equal to two right angles).
Subtract equal angles GFE and FEH. Therefore angle FEC = angle EFB. Q. E. D. 1.
Or the alternate-interior angles are equal when a straight line falls on two parallel lines.

(2)

Produce EF to I.
Then angle GFE = angle IFB (being vertical angles).
But GFE = FEH (alt. int. angles).
Therefore angle IFB = angle FEH (Ax. 1). Q. E. D. 2.
Or the exterior-interior angles on the same side of the transversal are equal when the lines are parallel.

(3)

But angle GFE + angle EFB is equal to two rt. angles.
But GFE equals FEH (alt. int. angles).
Therefore EFB + FEH = 2 rt. angles. Q. E. D. 3.
Or the sum of the interior angles on the same side of the transversal is equal to two right angles.

PROPOSITION V

If a straight line falling on two straight lines makes the alternate-interior angles equal to one another, or the exterior angle equal the opposite and interior angle, or the two interior angles on the same side of the transversal equal to two right angles, the lines are parallel.

(1)

Let AB and CD be two straight lines, and let EF be a third straight line making the alternate-interior angles AFE and DEF equal. Then the lines are parallel.
Proof.
For take any point G on AB, and draw EG. On the line CD in the direction ED take EH equal to FG, and join FH.
Then the angle FEH equals angle GFE (by hypothesis).
Side FE is common.
Line GF equals EH (by construction).
Therefore triangle GFE equals triangle FEH.
Therefore GE equals FH (corresponding sides of equal triangles).
Therefore, the lines AB and CD are parallel to one another (being equidistant at equidistant points). Q. E. D. 1.

(2)

If the opposite and interior angle is equal to the exterior angle, then the alternate-interior angles are also equal, since the exterior angle is equal to the opposite vertical angle, and hence the lines are parallel, from the previous proposition. Q. E. D.

(3)

If the sum of the interior angles on the same side of the transversal is equal to two right angles, the alternate-interior angles are equal and the lines are parallel.
For angles BFE + FEH = angles BFE + EFG (both equal 2 rt. angles). Subtract angle BFE from both, then angle GFE = angle FEH. Q. E. D.

THEOREM VI

(Prop. 32 of Euclid)

In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.

The proof that Euclid gives for this is now perfectly legitimate and valid, since we have proved all that is required for it. The difficulty with his proof is that it rests on his twenty-ninth, which in turn rests on the parallel postulate. But since we have proved the twenty-ninth in our fourth theorem without having had recourse to the fifth postulate but independently of it, this proposition can now be used to prove the thirty-second, and in this manner the proof offered by Euclid is complete and finally demonstrative.

Our whole method of proving the fifth postulate was to drop as useless the sixteenth theorem of Euclid, and change his method of proof as used in the twenty-seventh, twenty-eighth and twenty-ninth, thirty-third and thirty-fourth. The theorems we have substituted for these require nothing more than the sixteenth of Euclid, and hence can all be placed anywhere after the fifteenth in Euclid’s arrangement. All that we have used in this proof are the propositions, axioms, and postulates up to and including the fifteenth theorem, omitting the fifth postulate. We have again used the fifteenth concerning vertical angles; the fourth, concerning the equality of triangles, when the two sides and included angle of the one equal the two sides and included angle of the other, and equal sides opposite equal angles; the eighth, to show that when the sides of two triangles are equal the included angles are equal. Besides these we have used only axioms 1 and 2 and postulate 1. The whole proof rests on these, and no other principle is involved.

This is the sum total of the assumptions and propositions used in the proof of the parallel theory. Moreover, these propositions used in the proof in no way depend for their demonstration on the famous fifth postulate or on the definition of parallels.

Our proof is therefore absolute. First, it is not circular; that is, it does not assume, either overtly or virtually, anything which it is expected to prove. Secondly, it does not introduce any new proposition either implicitly or explicitly, as an axiom or postulate to be used in demonstrating the theorem. In the next place, it is a direct proof immediately applicable to the proposition to be demonstrated, and not a discussion of different possible theories, and a further attempt at reducing all but the chosen one to a contradiction, or by the argument known as reductio ad absurdum. It is moreover apodictic or necessary, and not a mere plausible proof, or one appealing to common sense and experience. Finally, it is a strict mathematical proof, since no non-mathematical element has been introduced, such as motion, etc. All other proofs brought forward have been objected to on all or most of these counts, and rightly so. It is not so much persuasion or conviction that is required, but proof according to the nature of the matter treated, and according to the proper method of proof in this special science. This we have done in the demonstration just submitted. The whole proof is built on the geometric relationship of the triangle to the parallelogram, and on the natural order that this should give rise to. The great mistake in Euclid’s method was that he did not recognize this fact.

Not only then is the geometry of Euclid solidly established without having recourse to the unproved assertion contained in the famous fifth postulate, but this postulate as well as the existence of parallel lines is strictly proved. We proved the existence of parallel lines as we defined them, in our propositions that we substituted for those of Euclid; and the fifth postulate is nothing more than a mere corollary from the proposition that the angles of a triangle are equal to two right angles, as we remarked before in connection with the first proof. But since, as we have seen, some doubt has been thrown on the definition of parallels as put forward by Euclid, we shall now treat this question separately.

References

1. Arist., Analyt., Post., I, 13.
2. Propositions and axioms used are numbered as in Euclid.