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The independence of the parallel postulate and the acute angles hypothesis on a two right-angled isosceles quadrilateral

We trace the development of arguments for the consistency of non-Euclidean geometries and for the independence of the parallel postulate, showing how the arguments become more rigorous as a formal conception of geometry is introduced. We use these philosophical views to explain why the certainty of Euclidean geometry was threatened by the development of what we regard as alternatives to it. Finally, we provide a model that creates the basis for proof and demonstration using natural language in order to study the acute angles hypothesis on a two right-angled isosceles quadrilateral, with a rectilinear summit side.


Introduction
Euclid's fifth postulate (called also the eleventh or twelfth axiom) states: "If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines if produced indefinitely meet on that side on which are the angles less than two right angles." The earliest commentators found fault with this statement as being not self-evident [1]. Euclid uses terms such as "indefinitely" and makes logical assumptions that had not been proven or stated. Thus the parallel postulate seemed less obvious than the others.
For two thousand years, many attempts were made to prove the parallel postulate using Euclid's first four postulates. Finally the mathematical society have proven the consistency of non-euclidean plane geometry and the independence of the euclidean parallel postulate; and extend that in stating that it would be impossible to prove Euclid's Parallel Postulate from the other assumptions made by him, since this would involve the denial of the Parallel Postulate of Bolyai and Lobatschewsky [2]. This conclusions mainly derived from Beltrami's (1868) seminal papers concerning the existence of non-Euclidean objects and by the arguments offered by Jules Houel in 1860-1870 for the unprovability of the parallel postulate and for the existence of non-Euclidean geometries. The final step towards rigorous consistency proofs is taken in the 1880s by Henri Poincare.
What Houel sees in Beltrami's putative demonstration that regions of pseudospheres are complete Bolyai-Lobatschewsky planes is the presentation of an instance in which all of the properties of the euclidean plane are present except for the property of unique parallelism [3]. A trivial example will make the principle evident. If we wish to show that the property of upright posture in mammals does not imply intelligence (in this sense of 'implication'), it is sufficient to show that there are some mammals in which upright posture is present without intelligence (kangaroos), even though there may be others in which upright posture occurs together with intelligence (humans). The application of the result here to the human case, e.g. if we want to say that upright posture in humans does not imply their intelligence, turns on referring to the same properties in the case of both humans and kangaroos (Scanlan 1988).
It's obvious that the independence of the euclidean parallel postulate is well formed, but a detailed looked at the writings will show that some claims may are overstated or misinterpreted. The most striking example of a non-logical use of the word "independent" is in Bolyai's title of his famous appendix, one of the founding texts of hyperbolic geometry, Absolute Geometry: Independent of the Truth or Falsity of Euclids Axiom XI (which can never be decided a priori). Bolyai does not, of course, claim that the Parallel Postulate is independent from the rest of Euclidean geometry in the sense required to show that the Parallel Postulate is unprovable (though he does claim that it cannot be proven). He means simply that he can prove some theorems of geometry without relying on either the Parallel Postulate or its negation [4].
Sometimes, it was simply stated that the attempt to prove the Parallel Postulate had gone on too long and that repeated lack of success shows that it is impossible to prove the Parallel Postulate. On the other hand the geometers working in order to support the existence of non-Euclidean objects were in the pursue of some very impressive results. That's why Beltrami's work had a profound impact on Houel. Houel immediately after he saw it, announced that Beltrami had shown that it is impossible to prove the Parallel Postulate. Strikingly, he made this announcement in eight different journals. In reality they wanted to strongly support and consolidate the initial hypothesis of the acute angle. The geometers had made just another logical assumption, believing that by finding a contradiction to Saccheri's quadrilateral would weaken the foundations of non-Euclidean geometries.
We can also realize that some statements wasn't very well formed or may were just some logical assumptions and we have to treat them with the same way we did with the logical assumptions that Euclid did. H. S. CARSLAW (1909) states: "How ever far the hyperbolic geometry were developed, no contradictory results could be obtained". Stating "How ever far" isn't it equivalent with the term "indefinitely" that Euclid uses?
Finally, let us consider the following: "for every diameter of γ let us imagine another point 'at infinity' such that all the Euclidean lines parallel in the Euclidean sense to this diameter meet in this point at infinity, just as railroad tracks appear to meet at horizon. These points at infinity will also be called ultra ideal" [5]. By imagining that the Euclidean lines meet to a point 'at infinity' we managed to show that these lines are parallel to the diameter γ, but are not parallel to each other. Nevertheless, in a specific area, by construction, we have lines that are parallel to each other due to the fact that they are parallel to a third line, something that we can not succeed without using the Euclidean sense of parallelism.
Finding a contradiction on the three hypotheses made by Saccheri does not mean that the non-euclidean geometries aren't consistent. It means that we can apply Euclid's proposition at these geometries without deconstruct them, because the independence of the Euclidean parallel postulate is bidirectional. It means that non-euclidean geometries are independent from the Euclidean parallel postulate and that the postulate, it self, is also independent from the geometries that exist in the Euclidean field. This concludes, that we can build relations without using it, but in the same time we can apply it anywhere we choose, without deconstructing our work. In reality Euclid did exactly the same thing, using the parallel axiom after the first 28 propositions. Thus was Euclid "vindicated" in an unexpected manner. Knowingly or not, the wise Greek had stated the case correctly, and only his followers had been at fault in their efforts for improvement [1].
The paper is organized in the following way, at section 2 we present the used Definitions, postulates and theorems. At section 3 we prove that the angles adjacent to the opposite sides of a quadrilateral, which are perpendicular to the base, are shorter or greater, if they are adjacent, respectively, to the greater or shorter sides. And they are equal if they adjacent to equal sides. At section 4 we prove that the bisector of the vertex angle of an isosceles triangle and the perpendicular bisector of the congruent sides of this triangle are intersecting. At section 5 we prove the rejection of the acute-angled and obtuse-angled quadrilaterals.

Used definitions, postulates and theorems
(1) The first four Euclid's postulates, the euclidean common notions and the theorems deduced by them. (2) The plane separation postulate.
(3) Theorems 1 and 2. We shall deal with them, in the course of our work.
3. Theorem 1 . The angles adjacent to the opposite sides of a quadrilateral, which are perpendicular to the base, are shorter or greater, if they are adjacent, respectively, to the greater or shorter sides. And they are equal if they adjacent to equal sides (corresponds to the Lemma I of G. Saccheri). proof: In quadrilateral ABΓ∆ (figure 1) we are given that angles A∆Γ and BΓ∆ are right angles and sides A∆ and BΓ are equal. Drawing the diagonals AΓ and B∆ two congruent triangles A∆Γ and BΓ∆ are formed, we prove this equality easily using the Side, Angle, Side (SAS) postulate. We may therefore conclude by the corresponding parts of the above congruent triangles, that the sides AΓ=B∆ and the angles A2=B2, ∆1=Γ1 . Since the angles ∆2 and Γ2 are complement of equal angles ∆1 and Γ1 they are themselves equal, that is, ∆2=Γ2. Using again the SAS postulate we also conclude that the triangles A∆B and BΓA are congruent, therefore by the corresponding parts of these triangles we conclude that the angles ∆AB and ΓBA are equal. . In quadrilateral ABΓ∆ (figure 2) we are given that angles A∆Γ and BΓ∆ are right angles, sides A∆ and BΓ are equal and angles ∆AB and ΓBA are equal too, as it has been proved above. On the line segment BΓ we take any point E between Γ and B, then by joining the points A and E the quadrilateral AEΓ∆ is formed. This new quadrilateral has as above, the angles A∆Γ and EΓ∆ right angles, but its opposite sides A∆ and EΓ are not equal now, that is, A∆ > EΓ. Since the angle AEΓ, as an exterior angle of the triangle ABE, is greater in measure than the interior remote angle ABE, the angles ABE and ∆AB are equal, and the angle ∆AB is greater in measure than the angle ∆AE, it follows that the angle AEΓ > ∆AE . Now the converse of the above statement. The greater in measure angle of a quadrilateral, which has the consecutive angles of its base right angles, is adjacent to its shorter side and vice versa.
In the given quadrilateral AEΓ∆ (figure 2) let angle AEΓ adjacent to the side EΓ be greater in measure than the angle ∆AE which is adjacent to the side A∆,  then according to the above argument, it follows that the side A∆ > EΓ, since if A∆ < EΓ or A∆=EΓ then it must be angle ∆AE > AEΓ or angle ∆AE=AEΓ. This last contradicts the original assumption that state angle AEΓ > ∆AE .
From now on, we shall call the above mentioned two right-angled isosceles quadrilaterals, which have the angles adjacent to the opposite and perpendicular to the base sides congruent, acute-angled quadrilaterals, right-angled quadrilaterals, or obtuse-angled quadrilaterals, if the pair of the congruent angles are, respectively, acute, right or obtuse angles.
3.1. The common property of the acute-angled, right-angled and obtuseangled quadrilaterals (the main property of Saccheri's quadrilaterals).
The perpendicular bisector to the base of the acute-angled, right-angled and obtuseangled quadrilaterals is also perpendicular bisector to the opposite the base side of them. proof: In acute-angled quadrilateral ABΓ∆ (figure 3) we are given that angles A∆Γ and BΓ∆ are right angles, ∆AB and ΓBA are equal acute angles and sides A∆=BΓ. We draw the line segments Z∆ and ZΓ, joining the midpoint Z of the side AB, with the endpoints ∆ and Γ of the side ∆Γ of the above quadrilateral ABΓ∆. We notice that the formed triangles ZA∆ and ZBΓ are equal, we can prove this equality using the side, angle, side postulate. We may therefore conclude by the corresponding parts of the above congruent triangles, that the angles Z3=Z4 and the sides Z∆=ZΓ. Since the triangle ∆ZΓ is isosceles it follows that the median ZE is also bisector of the vertex angle ∆ZΓ and perpendicular bisector to the base ∆Γ of the above triangle, hence the angles Z1 and Z2 are equal. Since the angles Z3, Z1, Z2 and Z4 are adjacent angles lie at the same straight line and the sums of the angles measures Z3+Z1 and Z2+Z4 are equal, it follows that the line segment ZE, which is perpendicular bisector to the base ∆Γ of the acute-angled quadrilateral ABΓ∆, is also perpendicular bisector to the opposite the base side, in which the acute angles of the quadrilateral are adjacent. By the same argument we prove that this property holds for the right-angled and obtuse-angled quadrilaterals too. Equivalent we can state that; in a two right-angled isosceles quadrilateral: the line segment which joins the midpoint of the base with the midpoint of the summit is perpendicular bisector to both of them.

Theorem 2
. The bisector of the vertex angle of an isosceles triangle and the perpendicular bisector of the congruent sides of this triangle are intersecting. Case A:. In the isosceles triangle ABΓ (figure 4) we are given that sides AB and BΓ are equal and the half of its base length AΓ equals to the length of the bisector B∆, of the vertex angle ABΓ, that is A∆=B∆. In this case the bisector of the vertex angle ABΓ and the perpendicular bisector to the side AB are intersecting at the midpoint ∆ of the base AΓ of the above triangle. proof: Since the side AB of the triangle ABΓ is now base for the isosceles triangle A∆B, it will be that the line segment E∆, which joins the midpoint of the line  segment AB and the vertex ∆ of the isosceles triangle A∆B, must be perpendicular bisector to the side AB . Case B:. In the isosceles triangle ABΓ (figure 5) we are given that sides AB and BΓ are equal and the half of its base length AΓ is shorter than the length of the altitude B∆, that is A∆ <B∆. In this case the bisector B∆ of the vertex angle ABΓ and the perpendicular bisectors to the congruent sides AB and BΓ are intersecting. proof: Since the side A∆ of the right triangle AB∆, is shorter in length than the side B∆, it results that the angle ∆AB is greater in measure than the angle AB∆. Furthermore by the equivalence of the plane separation postulate and Pasch's postulate [6] it follows that the perpendicular bisector ΠT except with the side AB has to intersect and an other side of the triangle AB∆. We suppose that the perpendicular bisector ΠT to the side AB of the triangle AB∆ also intersects and the side A∆ of the above triangle at a point M, since every point on the perpendicular bisector ΠT is equidistant from the endpoints of the line segment AB, it must be that the sides AM=MB and the angles BAM and ABM are equal, but this is impossible since the angle BAM=BA∆ is greater in measure than the angle AB∆ and angle AB∆ is greater in measure than the angle ABM. Therefore the perpendicular bisector ΠT except the side AB intersects always the bisector of the vertex angle ABΓ too. Case Γ: In the isosceles triangle AMΓ (figure 6) we are given that sides AM and MΓ are equal and A∆ > M∆, then the bisector of the vertex angle AMΓ and the perpendicular bisector of the side AM are intersecting. Proof: We are given that the line segment AΓ, and the perpendicular bisector of it(ΣZ) are intersecting at a point ∆. Using the point ∆ as a center and the line segment A∆, which is equal in length to the half of the line segment AΓ, as a radius we draw the circle(∆, A∆), this circle intersects the perpendicular bisector ΣZ at the points E and H. Using the point H as a vertex and the line segment AΓ as a base we form the isosceles triangle AHΓ, we notice that its altitude H∆ equals in length to the half of the base AΓ. Using the point of intersection E of the above circle(∆, A∆)with the perpendicular bisector ΣZ as a center and the cord of the same circle(∆, A∆) EA as a radius, we draw the circle(E, EA) this second circle intersects the perpendicular bisector ΣZ at the points Θ and N. Since in the circle(∆, A∆) the cord EA is shorter in length than the diameter EH, it follows that the point N lies between the points H and ∆, and the altitude of the formed isosceles triangle ANΓ is shorter in length than the half of the common base AΓ. By repeating the above used procedure we can form isosceles triangles, which have the given line segment AΓ as a base and an altitude in length, in comparison with the half of its base AΓ, as short as we want. This last takes place, since by the repetition of the above referred procedure, the arc of the given cord AΓ continuously diminished tend to coincide with it. Since the above described isosceles triangles are inscribed in circles, it follows that the perpendicular bisectors to their equal sides intersect the common bisector of the vertex angles of them ΣZ. . Let AMΓ be any isosceles triangle in which the length of the altitude M∆ is shorter than the half of the base length AΓ. We remind here that for any isosceles triangle AMΓ, with an altitude M∆ we can form another inscribed isosceles triangle ANΓ with the same base, collinear vertex angle bisector and with an altitude in length N∆ < M∆. Then the perpendicular bisectors to its equal sides will intersect the bisector ΣZ, of the vertex angle AMΓ, in respect of the following arguments. Given that the line segment θM is greater in length than θN and θN is greater than θA, since θN and θA are, respectively, diameter and cord in the same circle (E,EA) it implys that θM > θA. From this we can conclude that in the triangle AθM the angle θAM is greater in measure than the angle θMA. This last ensure that the perpendicular bisectors to the equal sides of the isosceles triangle AMΓ will intersect the bisector of the vertex angle of it. Since the opposite assumption, which states that the perpendicular bisector to the side AM will intersect the side Aθ in the triangle AθM, given the Pasch's postulate, implys that the angle θMA is greater in measure than the angle θAM which is a contradiction.

5.
The rejection of the acute-angled and obtuse-angled quadrilaterals . In the acute-angled quadrilateral AEZ∆ (figure 7) we are given that the line segment ΓM is perpendicular bisector to the sides ∆Z and AE, because it joins the midpoints of them, see the common property of the acute-angled, right-angled and obtuse-angled quadrilaterals. Since the right angle AMΓ=EMΓ is greater in measure than the congruent acute angles ∆AM and ZEM it results, according to the theorem 1, that in the quadrilaterals AMΓ∆ and MEZΓ the sides A∆=EZ > MΓ. Following, we extend the line segment ΓM, we take on it a point B, so that A∆=BΓ=EZ, we draw the line segments AB and BE and then we notice that the formed triangle ABE is an isosceles triangle, since the line segment BΓ, which now is bisector of the angle ABE, is perpendicular bisector to the side AE. By repeating the above procedure we take the line segment OΠ perpendicular bisector to the sides ∆Γ and AB of the acuteangled quadrilateral ABΓ∆, the ABΓ∆ is an acute-angled quadrilateral, since the angles A∆Γ and BΓ∆ are right angles, the sides A∆ and BΓ are equal, and the angles BA∆ and ABΓ are congruent acute angles, because these are adjacent to the equal sides A∆ and BΓ, and the angle ABM=ABΓ is a non right angle of the right triangle ABM. Similarly we take the line segment ΣT perpendicular bisector to the sides ΓZ and BE, for the same mentioned above reasons. From the above analysis we can conclude that the perpendicular bisectors OΠ and ΣT to the congruent sides of the isosceles triangle ABE, as well the bisector of the vertex angle ABE are not intersected, since they are perpendiculars to the same side ∆Z, and the perpendicular to a line through a point not on the line is unique. But this conclusion contradicts to the results of the theorem 2. . Using the same procedure, and taking into account that in the obtuse-angled quadrilateral AEZ∆ (figure 8) the perpendicular bisector ΓM, to the base ∆Z and to the summit AE, is longer than the congruent sides A∆ and EZ, because the equal obtuse angles ∆AM and ZEM are greater in measure than the right angles AMΓ and EMΓ, we conclude again that the perpendicular bisectors to the equal sides of the isosceles triangle ABE, as well the bisector of the vertex angle ABE are not intersected. But this conclusion also contradicts to the results of the theorem 2.

Conclusions
. Since from the above conclusions we have to reject the hypotheses that the congruent angles adjacent to the equal opposite sides of the quadrilaterals, in which their base angles are right angles, are acute angles or obtuse angles, since these contradicts the results of the theorem 2, we have to accept that the quadrilaterals in which their base angles are right angles and their opposite sides are equal to each other and perpendicular to the base are only rectangle quadrilaterals.