# What equidistant from one another” and “parallel

Among them were “parallel lines are everywhere equidistant from one another” and “parallel lines are line having the same direction from a given line.” But these early definitions were sometimes vague or contradictory. Euclid tried to overcome these difficulties by his definitions, “parallel lines are straight lines which, being in the same plane and being produced indefinitely in both directions do not meet another in either direction.” Euclid’s fifth postulate 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 line produce indefinitely meet on that side on which are the angles less than two right angles.” Euclid’s assumptions fall into one of the two categories. First, the set of “self-evident” facts concerning plane figures.

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An example of this assumption is that “a straight line is the shortest distance between two points.” Second, it deals with the concept beyond realm of actual experience. For instance, Euclid stated that “a straight line must continue undeviatingly in either direction without end and without finite length.” Since, it is impossible to experience things indefinitely far off, anything that is said about events, there is speculation, not self-evident truth. The fifth postulate falls into this latter category. The complicated nature of the fifth postulate led numerous mathematicians to believe that it could be proved using the remaining postulates. Many mathematicians attempted to prove Euclid’s fifth postulate.

Among those who attempted a proof of the parallel postulate was Proclus. The statement Proclus proves instead of the parallel postulate is “Given ? + ? < 2d, prove that straight line y? and y?? meet at a certain point C. In his proof, Proclus draws a straight line, y???through a given point parallel to y?. Then taking a point B on y?? he draws a perpendicular line to y??? from it. From this, he reasons that since the distance from y??? increases without limit as the distance between a and B grows and the distance between y? and y?? is constant that there must be a point C on y? and y??, and it is the point where y? and y??meet, thus completing his proof. However, as with most of the other alternatives to the parallel postulates, this one had faults. It is observed by Pogorelov that the parallel straight lines proofs rely on are not explicitly contained in the other postulates or axioms and therefore cannot be deduced from them.

Another person who attempted a proof to parallel postulate was John Wallis. However, instead of proving the theorem directly with neutral geometry, he proposed a new axiom. This postulate expressed the idea was that one could either magnify or shrink a triangle as much as one likes without distortion. Using this, Wallis proves the parallel postulate as follows: He begins with two straight lines making, with third infinite straight line, two interior angles, less than two right angles. Then he slides one angle down the line AF until it reaches a designated position ??, cutting the first line at ?. Then using his first postulate, he claims that the two triangles ?C ? and ACP are similar, thus showing that AB and CD meet at point P, and proving the theorem.

However, this too had a fault. In fact, the original postulate that he based the proof on was logically equivalent to Euclid’s fifth postulate. Therefore, he had assumed what he was trying to prove, which make his proof invalid. Another group to prove on Euclid’s parallel postulate was the Medieval Islam. One mathematician from this time who contributed to clarification of Euclid’s parallel postulate was Abu’ Sli Ibn al-Haytham.

He begins his proofs by first addressing Euclid’s definition of parallel lines. He states that “parallel straight lines are coplanar lines such that if produced indefinitely in both directions they do not intersect in either direction” which is phrased which is phrased so that all the ;lines “involved will be segments of finite length. He then begins his proof. First, he begins with a given vertical line AB. Then from an endpoint A he constructs a line AC, forming a right angle with AB. Similarly, he constructs a line BD, meeting at endpoint D.

In this figure al-Haytham claims that CD is equal to AB. To prove this, he uses a proof by contradiction. He first supposes that CD is greater than AB. He then extends the line CA through A to form AE. Similarly, he forms the line BF through B. Next al-Hay ham cuts the line AE so that AE is equal in length to AC, after which he drops a perpendicular from E to F.

Finally he draws the lines CB and BE. To continue his proof, al-Haytham needs to show that line CD is equal to EF, and that both are greater than AB. Using what we now refer to as the Side-Angle-Side Theorem, he says that since CA is equal to AE, angle CAB is equal to angle EAB (right angles), and the side AB is common, therefore the triangles CAB and EAB must also be equal. Thus line CB is equal to EB and the two remaining angles must also be equal. al-Haytham continues that angle CBA and angle EBA are equal, and since angles ABD and ABF are equal, therefore the angles CBD and EBF must also be equal. Next, by what we now refer to as the Side-Angle-Side Theorem, he claims that since the angles CDB and EFB are equal and sides DB and BF are equal; therefore the triangles CDB and EFB must be equal. Therefore, CD and EF are also equal.

Then, since CD is greater than AB (by assumption) EF must also be greater than AB. Next in al-Haytham's proof, he says to imagine EF moving along FB so that the angle EFB remains a right angle throughout the motion, with EF remaining perpendicular to FB. Then when point F coincides with point B, line EF will be "superposed" onto AB.

But he claims that since the magnitude of EF is greater than that of AB, point E will lie outside AB (on the same side with A). Thus at this point EF is equal to HB. Next al-Hay ham slides line BH along BD. If in this process point B coincides with point D then BH will be “superposed; on DC (because angles HBF and CDB are equal).

Then since BH = EF = CD, al-Haytham claims that H coincides with C. Thus al-Haytham has showed that if EF is put in motion along FD, then the points E and F will coincide with C and D, respectively. Next he notes that if any straight line moves in this way then it's ends will describe a straight line. Thus the point E describes the straight line EHC. al-Haytham concludes that since H does not lie on AB it cannot coincide with point a and therefore there must exist a surface bound by two straight lines which he finds to be absurd ;, therefore proving that CD is neither greater nor less than A.

Thus al-Haytham has showed that CD and all other perpendicular lines dropped from AC to BD are equal to AB.In conclusion, throughout the past 2300 years of mathematical history many mathematicians from all around the world have unsuccessfully been trying to prove Euclid's parallel postulate. Although these attempted proofs did not lead to the desired result, they did play a part in the development of geometry, enriching it with new theorems that were not based on the fifth postulate, as well as leading to the construction of a new geometry, Non-Euclidean geometry, not based on the parallel postulate.

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