Download Math 324 - Corey Foote

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Corey Foote
Dr. Kent
Honors Geometry
20 April 2011
Neutral (Absolute) Geometry: An Exploration of its History and Implications
Most of us are familiar with the geometry that we learned in basic high school and
introductory level college courses. This is called Euclidian geometry. Euclidian geometry is
based on Euclid’s five postulates; in other words, all geometry proofs can be started because
these first five postulates are assumed. Many geometers after Euclid’s time thought the first four
postulates were self-evident; however, they felt the fifth one could not just be a simple
assumption. The geometry that does not use Euclid’s fifth postulate as an assumption is known
as neutral, or absolute, geometry.
Before delving into neutral geometry, we must first know Euclid’s first four postulates.
He says the following: “Let the following be postulated: (1) To draw a straight line from any
point to any point. (2) To produce a finite straight line continuously in a straight line. (3) To
describe a circle with any center and distance. (4) That all right angles are equal to one another”
(Maita and Rowling 592). These are fairly straightforward. The first postulate assumes that you
can draw a straight line between two points. The second postulate says you can take a finite line
segment and extend it arbitrarily far. The third postulate states that a circle can be made if you
have a center and a radius. The fourth postulate simply says that right angles are equal to each
other.
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Now it is time to discuss Euclid’s fifth postulate. It is very easy to see just why many
mathematicians deem it too complicated to be an assumption. It states, “That 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” (Fitzpatrick 599). Of course, this is a very complicated version of the fifth
postulate. Martin Isaacs of the University of Wisconsin, Madison, states a similar, yet easier-tounderstand, definition of the parallel postulate: “given a line and a point on that line, there exists
one and only one line through the given point parallel to that line” (Isaacs 4). This version of the
parallel postulate is known as Playfair’s Axiom. Clearly, the fifth postulate seems very
complicated; it seems more like a theorem than an assumption. Many mathematicians held this
opinion, and they tried to prove it. Their attempts failed; Euclid, however, did prove the converse
of the parallel postulate (Levine and Rowling, 592).
Euclid’s Proposition I.17 is actually the converse of the parallel line postulate
(Bogomolny). Let us prove the converse of this postulate. Proposition I.17 states the following:
“In any triangle two angles taken together in any manner are less than two right angles” (Euclid,
281). Please note that fleshed-out versions of all proofs will be provided. Also, follow Figure 1
below.
Figure 1
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Proof: First, let ABC be a triangle. Extend BC to D. Since ∠ACD is an exterior angle of
∆ABC, it is greater than ∠ABC. Next, add ∠ACB to ∠ACD and ∠ABC. It follows that ∠ACB
and ∠ACD is greater than ∠BCA and ∠ABC. Also, ∠ACB and ∠ACD are supplementary angles,
and so must add up to the measure of two right angles, or 180°. It follows that ∠BCA and ∠ABC
are less than two right angles, as they are less than the sum of ∠ACB and ∠ACD. By using
similar reasoning, ∠BAC and ∠ACB, along with ∠CAB and ∠ABC, are less than right angles.
Q.E.D.
Without assuming a parallel postulate, we have neutral (sometimes called “absolute”)
geometry. This means that mathematicians take a neutral position about the parallel postulate
(Greenberg, 161). Much of what we know from a basic study of geometry is also in neutral
geometry, including the concepts of acute, obtuse, and right angles. Also, the topics of
complementary, supplementary, and adjacent angles, as well as triangles and quadrilaterals, are
elements of neutral geometry (Wallace and West, 66). Students learn these topics as far back as
elementary school and high school geometry and are thus introduced to the basics of neutral
geometry in their early years.
To show some examples, let it be known that Euclid did not use the parallel postulate for
his first twenty-eight propositions (Greenberg, 162). Let us take a look at his fourth proposition:
“If two triangles have the two sides equal to two sides respectively, and have the angles
contained by the equal straight lines equal, they will also have the base angle equal to the base,
the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining
angles respectively, namely those which the equal sides subtend” (Euclid, 247). This proposition
deals with triangle congruency, which is as basic as an entry level college geometry course can
get. This is the side-angle-side congruency theorem.
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Look at Figure 2 and Figure 3 as you follow along with the proof. As one can easily see,
this proposition does not rely on the use of Euclid’s fifth postulate.
Figure 2
Figure 3
Proof: First, let AB and AC be equal to DE and DF, respectively. Also, since we are
assuming that angles contained by equal straight lines are equal, we can let ∠BAC = ∠EDF.
Proceed to line up the triangles in such a way so that B will coincide with E, therefore having
base BC coincide with base EF. Since the triangles coincide, the corresponding angles and bases
are equal. Hence, ∠ABC = ∠DEF, ∠ACB = ∠DFE, and BC = EF. Q.E.D.
Let us now take a look at one of Euclid’s propositions that requires the use of the
controversial fifth postulate, the twenty-ninth proposition. This is the very first proposition in
which Euclid relies on the parallel line postulate. This states, “A straight line falling on parallel
straight lines makes the alternate angles equal, the exterior angle equal to the interior and
opposite angle, and the interior angles on the same side equal to two right angles” (Euclid, 311).
The following proposition assumes parallel lines. Look at Figure 4 to see this proof illustrated.
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Figure 4
Proof: Let the straight line EF fall on the parallel and straight lines, AB and CD. The
alternate interior angles must be equal. Hence, ∠AGH =∠DHG and ∠BGH = ∠CHG. The
exterior angles must also be equal. Therefore, ∠EGB = ∠FHC and ∠EGA = ∠FHD. Also,
interior angles on the same side of the transversal must add up to 180°. Hence, ∠BGH and
∠DHG, as well as ∠AGH and ∠CHG are supplementary.
To show these are results of a transversal falling on two parallel lines, we must use proof
by contradiction. For example, if ∠AGH is unequal to ∠GHD, two angles that, by definition,
should be equal, one of them clearly has to be greater. Make ∠AGH greater than ∠GHD. Proceed
to add ∠BGH to each of these angles. Hence, ∠AGH and ∠BGH are greater than ∠BGH and
∠GHD. However, ∠AGH and ∠BGH are supposed to be equal to 180°. By looking at Figure 4,
one can see that they are supplementary angles, since ∠AGB = 180° and ∠AGB = ∠AGH
+∠BGH. Because ∠AGH and ∠BGH are supposed to equal 180°, ∠BGH and ∠GHD must be
less than 180°, as shown above.
Now we see Euclid’s fifth postulate in this proposition. The second part of Euclid’s fifth
postulate states, “But straight lines produced indefinitely from angles less than two right angles
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meet.” Because AB and CD are produced by ∠BGH and ∠GHD, they must meet. However, it is
stated in the hypothesis that they are equal. Therefore, ∠AGH must equal ∠GHD.
Now consider these angles that are supposed to be equal: ∠AGH = ∠EGB and ∠EGB =
∠GHD. Add ∠BGH to each. Hence, ∠EGB and ∠BGH are equal to ∠BGH and ∠GHD. Since
∠EGB and ∠BGH are supplementary, ∠BGH and ∠GHD are also supplementary, since they are
equal. Q.E.D.
Now we will take a look at Hilbert’s parallel postulate, which is equivalent to Euclid’s
parallel postulate. This states: “For every line l and every point P not lying on l there is at most
one line m through P such that m is parallel to l.” As you can clearly see, this is not as strong as
Euclid’s fifth postulate. Euclid’s fifth postulate, as previously mentioned, implies that there is “at
least one line through P that is parallel to l (Greenberg, 138). Hilbert’s postulate states that there
is a most one line. This is very limited.
Now that we know what Hilbert’s parallel postulate states, we can prove its equivalency
to Euclid’s parallel postulate. However, we are not able to prove both of these postulates; we can
only prove one if we assume the other.
Theorem 1: Euclid’s fifth postulate is equivalent to Hilbert’s Euclidian parallel postulate.
(Greenberg, 173).
Before we can prove Theorem 1, we must have a list of the required theorems and
corollaries
Theorem 2 (Alternate Interior Angle): In any Hilbert plane, if two lines cut by a
transversal have a pair of congruent alternate interior angles with respect to that transversal, then
the two lines are parallel (Greenberg, 162).
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Corollary 3 (To Theorem 2): Two lines perpendicular to the same line are parallel
(Greenberg, 163).
Theorem 4 (Exterior Angle Theorem): In any Hilbert plane, an exterior angle of a
triangle is greater than either remote interior angle (Greenberg, 164).
Theorem 5: If ∠B is supplementary to ∠A, then (∠A) ° + (∠B) ° = 180°.
Here is the proof of Theorem 1with a detailed explanation (Greenberg, 170).
Figure 5
Proof: First, let’s assume Hilbert’s parallel postulate. Using Figure 5 above, which
represents Euclid’s postulate, assume ∠1 + ∠2 < 180°. Also, ∠1 + ∠3 = 180° by Theorem 5. It
follows that ∠2 < 180° - ∠1 = ∠3. Now consider ray B′C′. This ray is unique in the fact that ∠3
and ∠C′B′B are alternate interior angles. By Theorem 2, B′C′ is parallel to l. Also, m ≠ line B′C′,
so m meets l. This is Hilbert’s postulate. In order to prove that m meets l on the same side of t as
C′, assume they meet at an arbitrary point A on the opposite side of t. If you draw ∆ABB′, ∠2 is
an exterior angle; however, it is smaller than ∠3, even though ∠3 is a remote interior angle. This
contradicts Theorem 4, which is Hilbert’s postulate. Hence, they must meet on the other side of
the transversal, which proves Hilbert’s postulate on Euclid’s situation. Thus, Euclid’s postulate is
proven when you first assume Hilbert’s postulate.
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Figure 6
Now, let’s assume Euclid’s parallel postulate. See Figure 6 above, which represents
Hilbert’s postulate. Let t be perpendicular to l through P, and also let m be perpendicular to t
through P. By Corollary 3, m ∥ l. Now, draw a random line, n through P. In order to prove
Hilbert’s postulate, we must prove that n meets l. Next, look at ∠1. Clearly, this angle is an acute
angle. Now add ∠PQR to this, which is a right angle. This sum is obviously less than 180°.
Euclid’s postulate is therefore satisfied on Hilbert’s situation. Thus, Hilbert’s situation is proven
when Euclid’s postulate is assumed. Q.E.D.
There are six consequences that are equivalent to the parallel postulate. “(1) Through a
given point only one parallel can be drawn to a given line. (2) There exists at least one triangle in
which the sum of the measures of the interior angles is 180°. (3) There exists a pair of similar,
but not congruent, triangles. (4) There exists a pair of straight lines that are everywhere
equidistant from one another. (5) Every triangle can be circumscribed. (6) The sum of the
measures of the interior angles of a triangle is the same for all triangles” (Wallace and West, 3435).
Wallace and West point out that many mathematicians have tried to prove that the fifth
postulate was a consequence of the first four postulates. They describe a way to prove the fifth
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postulate if we could prove one of the six consequences of the parallel postulate listed above.
However, one would have to work backwards to do this (Wallace and West, 35).
Theorem 6: Assuming the parallel line postulate, the sum of the measures of the interior
angles of a triangle is 180°.
Figure 7
Proof: Look at Figure 7 above. Let x be the sum of the interior angles of a triangle. The
total measures of the left-hand triangle is therefore ∠3 + ∠4 + ∠5 = x. The total measures of the
right-hand triangle is ∠1 + ∠2 + ∠6 = x. By using basic middle school algebra, you can combine
the equations as such: ∠3 + ∠4 + ∠5 + ∠1 + ∠2 + ∠6 = 2x. Now, set ∠5 + ∠6 = 180°. You can
easily do this because these two angles are supplementary to one another. By definition, their
sums must equal 180°. Hence, x = 180°. Q.E.D.
Why is x = 180°? By using the second consequence of the parallel postulate, you can
assume that ∠3 + ∠4 + ∠5 and ∠1 + ∠2 + ∠6 each add up to 180°. Although you cannot see the
parallel postulate used directly, you can see one of the consequences has been used to prove it.
Marvin Greenberg shows us an alternate way to prove this without relying directly on the
Euclidian parallel postulate. We need the following theorem before we can start this proof.
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Theorem 7: Hilbert’s Euclidian parallel postulate is equivalent to the converse of
alternate interior angle theorem (Greenberg, 175).
Now8let us
Figure
Theorem 8: In any Hilbert plane, Hilbert’s Euclidian postulate implies that for every
triangle ∆ABC, ∠A + ∠B + ∠C = 180°. We must assume Hilbert’s parallel postulate in order to
prove that the sum of ∆ABC and the sum of every triangle in general, is 180°.
Proof: Assume Hilbert’s parallel postulate. Also, look at Figure 8 above. By Corollary 3,
there is a line through B that is parallel to AC. Extend BC and BA through this line. By Theorem
7, the alternate interior angles with respect to these transversals are equal. Hence, ∠ABE
=∠ABE and ∠CBD =∠BCA. It follows that ∠ABC plus the two exterior angles equal 180°.
Hence, because those same exterior angle measures are inside the triangle, ∆ABC = 180°. Q.E.D.
As you can see, to prove certain things that normally rely on the parallel postulate, one
must either use consequences of the parallel postulate or Hilbert’s Euclidian postulate, which, as
already stated, is equivalent to Euclid’s parallel postulate.
In the article “The Pythagorean Theorem is Equivalent to the Parallel Postulate,” Scott
Brodie gives us the equivalencies of the parallel postulate, such as the Pythagorean Theorem.
Assuming the Pythagorean Theorem proves that there are 180° in a triangle, etc. The last
theorem then implies the parallel postulate. Look at Figure 9 to see just how each theorem
implies the next theorem and how they are all related.
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Figure 9
I.) Through a given
point, only one line
can be drawn
parallel to a given
line.
IV.) There exists an
isosceles right
triangle whose three
angles sum to two
right angles.
VI.) The three
angles of any
right triangle
sum to two right
angles.
II.) In a right triangle, the
square of the hypotenuse
equals the sum of the
squares on the other two
sides.
III.) There exists
some triangle
whose three
angles sum to
two right angles
V.) There exists
arbitrarily large
isosceles right
triangles whose
angles sum to two
right angles
I.) Through a
given point,
only one line
can be drawn
parallel to a
given line.
Let us take a look at the first equivalency. This is proof shows how the parallel line
postulate implies the Pythagorean Theorem. The proof of this is actually Euclid I.47, which
simply proves that a2 + b2 + c2 in right triangles. Follow Figure 10 as we go through this proof.
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Figure 10
Theorem 9: The parallel line postulate implies the Pythagorean Theorem.
Proof: First, assume the parallel line postulate. Let ∆ABC be a right triangle. The right
angle of this triangle is ∠BAC. Prove that in right-angled triangles, the square on the side
subtending the right angle is equal to the squares on the sides containing the right angles. In the
case of ∆ABC, prove the square of BC be equal to the sum of the squares of BA and AC.
Using side BC, create square BDEC below ∆ABC. Similarly, using side AB, create
square AGFB and, using side AC, create square AHKC. Draw AL parallel to CE and BD. (This
is a direct use of the parallel postulate.) Also draw AD and FC.
We already know ∠BAC is a right angle. Also, ∠BAG is a right angle because it is part
of square AGFB and, by definition, all angles in a square are 90°. Now consider BA. When
looking at AC and AG, notice that they do not lie on the same side as BA. These two segments,
AC and AG, make the two adjacent angles, ∠BAG and ∠BAC, equal to two right angles. Hence,
CA is a straight line with AG. Without a loss of generality, BA is a straight line with AH.
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Next, consider how ∠DBC = ∠FBA = 90°. Add ∠ABC to each of the two angles. Clearly,
when looking at Figure 8, ∠DBA = ∠FBC. We know DB = BC and FB = BA. We also know
that AB = FB and BD = BC, along with ∠ABD = ∠FBC. By SAS, ∆FBD ≅ ∆ABC. Hence, AD =
FC because they are corresponding sides of two congruent triangles.
Now consider parallelograms BMLD and ∆ABD. Parallelogram BMLD is double ∆ABD
because they both lie between parallels BD and AL, and they share the base BD. Recall that a
triangle is half of a parallelogram assuming these conditions. Without a loss of generality, square
AGFB is double ∆FBC. Also remember that a square is a parallelogram, so this works the same
as the previous case. It follows that KBMLD = KAGFB. Without a loss of generality, KCMLE =
KHKCA.
Since we want to prove that KHKCA = KAGFB + KBCED, it is first necessary do a
substitution. We know that KAGFB = KBMLD from earlier. We also know that that KBCED = KBMLD+
KMCEL. We can substitute and say KBCED = KAGFB + KMCEL. Now, since we know KHKCA =
KMCEL, we can do another substitution. Therefore, KBCED = KAGFB + KHKCA. Now reconsider
∆ABC. Look at the sides of this triangle. Clearly, BC is a part of BCED, AB is a part of AGFB,
and AC is a part of CKHA. Squaring these sides will give you the areas of the respective squares,
which allows us to use this formula: (AB) 2 + (AC) 2 = (BC) 2. Q.E.D.
Now we shall see how the second statement implies the third statement. We must first
look at and understand Theorem 10. Then, we must look at three lemmas that A.M. Legendre
wrote. Lemma 11 is used to prove Lemma 12 using a proof by contradiction technique. Lemma
12 also uses proof by contradiction to prove Lemma 13. Then we will use Lemma 13 to help us
prove Theorem 14.
Theorem 10 (pons asinorum): Base angles of an isosceles triangle are equal (Isaacs, 8).
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Lemma 11: In any triangle, the sum of any two angles must be less than two right angles
(Brodie).
Lemma 12: In any triangle, the sum of the three angles is less than or equal to two right
angles (Brodie).
Lemma 13: If the sum of the angles of a triangle equal two right angles, and if a segment
is drawn from one vertex to the opposite side, so as to divide the triangle into two smaller
triangles, then the angle sum of each of the smaller triangles is also two right angles (Brodie).
Figure 11
Proof: Draw AX on Figure 11 above, but do not make it a midpoint of side BC. We
know that the sum of the angles of ∆ABC is equal to two right angles. Take ∆ABX and denote
the sum of its angles measures by A1. Similarly, denote the sum of the angle measures of ∆ACX
as A2. Also, let the sum of the angle measures of ∆ABC be denoted by A3. If the following
statement, A3 = A1 + A2 = 180° is true and A2 < 180°, then A1 must be greater than 180°. This
contradicts Lemma 12, which states that the sum of the angles in a triangle must be less than or
equal to 180°. The sum of the angles in ∆ABX is greater than the sum of two right angles.
Hence, by proof by contradiction, we prove our argument. Q.E.D.
Theorem 14: If there exists one triangle whose three angles sum to two right angles, then
there exists an isosceles right triangle whose angle sum equals two right angles (Brodie).
Foote 15
Figure 12
Proof: Look at Figure 12 above. Assume ∆ABC where BC is the shortest side of the
triangle. Place a point D on side AB such that BC = BD. Then draw CD. Now ∆DBC is
isosceles. By the pons asinorum, ∠CDB = ∠CBD. Now draw CE such that it is the altitude of
∆DCB. Hence, ∠CEB = 90°. By Lemma 13, ∆CEB is a right isosceles triangle whose angle
measures sum to two right angles. In conclusion, there is at least one right isosceles triangle
whose angle measures add up to two right angles. Q.E.D.
Now let us look at how the third statement implies the fourth statement. Look at Figure
12 on the next page.
Figure 13
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Theorem 15: If there exists an isosceles right triangle whose angles sum to two right
angles, then there exists an arbitrarily large isosceles right triangle whose angles sum to two right
angles (Brodie).
Proof: Let us take the right isosceles ∆CEB from the previous theorem. Duplicate this
triangle and call it ∆C'E'B'. Line these two triangles up letting the hypotenuses of each triangle
coincide. This forms quadrilateral CE'C'E. (This can be labeled in multiple ways.)
The angles in ∆CEB must add up to 180° and ∠E = 90°. Therefore, ∠C = ∠B = 45°
because the pons asinorum states that base angles of an isosceles triangle must be equal. Since
∆C'E'B' is just a mirror image of ∆CEB, we can conclude that ∠C = ∠B = ∠B' = ∠C ' = 45°. It
follows that ∠C + ∠B' = 90° and ∠B + ∠C' = 90°. Also, all four sides are equal since we are
combining two isosceles triangles. By definition, quadrilateral CE'C'E is a square.
Next, increase the lengths of the sides by three times the current length to form a new
square. Draw a diagonal from E to E'. Clearly, the two triangles that are formed by drawing the
diagonal are right isosceles triangles, since they contain a right angle and two equal sides. Note
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that these sides are still equal because each side length was multiplied by three. Hence, there is a
larger right isosceles triangle whose angle measures sum to two right angles. Q.E.D.
Now we shall see how the fifth statement implies the sixth statement.
Theorem 16: If the angles of an arbitrarily large isosceles right triangle sum to two right
angles, then the angles of every right triangle sum to two right angles (Brodie).
Figure 14
Proof: Let ∆E′BE be an arbitrarily large isosceles right triangle whose angle measures
sum to two right angles. Now place point X anywhere on E′B in Figure 13. Similarly, place Y
anywhere on EB. Do not place these points equidistant from B. We do this because we are trying
to show that right triangles have three angles whose sums are equal to 180°. If we make these
points equidistant, we are proving that right isosceles triangles have the sum of their angle
measures equal to two right angles. We have already proven this fact. Connect the two points to
form ∆XYB.
Next, extend point Y to E′. Since YX divides ∆E′YB into two triangles, ∆E′YX and
∆XYB, by Lemma 12, both triangles have the sum of their angles equal to two right angles.
Foote 18
Clearly, ∆XYB is a right triangle, so the angles of every right triangle sum to two right angles.
Q.E.D.
We are almost finished proving the equivalency chart. Our final task is to show how the
sixth statement, which we have just proven, implies the parallel postulate. Follow Figure 14
below.
Figure 15
Theorem 17: If the three angles of any right triangle sum to two right angles, then,
through a given point, only one line can be drawn parallel to a given line.
Proof: Draw line l, as in the figure above. Place point P on line r and draw QP
perpendicular to line l. Now draw line r in such a way that it goes through P but does not
intersect l. If it were to intersect l, we would have a triangle whose angle sums measure more
than two right angles. This clearly contradicts Lemma 11.
Next, draw a line s, which goes through P and is also parallel to l. Choose a point T on
this line such that ∠TPQ < 90°. Let 90° - ∠TPQ = ∠TPV = a. Next, draw a line, n, through P that
intersects l at point U in such a way that ∠PUQ < ∠TPQ. Also place V on r on the same side of
PQ as U.
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Since ∠PQU is a right angle, by Theorem 16, ∠PUQ and ∠QPU must add up to 90°. Also
note that ∠QPV is a right angle. Therefore, the sum of ∠QPV, ∠QPU and ∠UPV, must equal a
right angle. Since, both sums are equal to 90°, ∠PUQ + ∠QPU = ∠QPU + ∠UPV. Thus, ∠PUQ =
∠UPV. Recall that ∠PUQ < ∠TPQ. It follows that ∠UPV < a = ∠TPV. Because of this, PT must
lie on the interior of ∠QPU. Since U intersects l, PT will intersect l at some point. This proves
that through a given point, only one line can be drawn parallel to a given line. Q.E.D.
The parallel line postulate is an essential part of Euclidian geometry. Although it seems
too basic to be considered a postulate, it has never been proven outright. It can only be proven if
one assumes something else (other than the four postulates). It is here that we have the branch of
Euclidian Geometry knows as neutral, or absolute, geometry.
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Works Cited
Bogomolny, Alexander. “Euclid’s Fifth Postulate.” Interactive Mathematics Miscellany and
Puzzles. 1996-2011. Cut The Knot. <http://www.cut-the-knot.org/MailNotification
Page.shtml>.
Brodie, Scott E. “The Pythagorean Theorem is Equivalent to the Parallel Postulate.” Interactive
Mathematics Miscellany and Puzzles. 1996-2011. Cut The Knot. <http://www.cut-theknot.org/MailNotificationPage.shtml>.
Euclid. Thirteen Books of Euclid’s Elements: Vol. 1. Translation from the Text of
Heiburg. New York: Dover Publications, 1956.
Fitzpatrick, Sister Mary of Mercy. “Chapter 91: Saccheri, Forerunner of Non-Euclidian
Geometry.” Swetz 597-605.
Greenberg, Marvin J. “Chapter 4: Neutral Geometry.” Euclidian and Non-Euclidian
Geometers. New York: W.H. Freeman and Company, 2008. 161-207.
Isaacs, Martin. Geometry for College Students. Providence: American Mathematical
Society, 2009.
Levine, Maita and Rolwing, Raymond H. “Chapter 90: The Parallel Postulate.” Swetz
592-596.
Swetz, Frank J, ed. From Five Fingers to Infinity. Chicago and La Salle, Illinois: Open Court
Publishing Company, 1994.
Wallace, Edward C. and Stephen F. West. Roads to Geometry. Englewood Cliffs, New Jersey:
Prentice Hall, 1992.