Non MATH 6118
Popular in Course
Popular in Mathematics (M)
This 118 page Class Notes was uploaded by Mrs. Dangelo Fahey on Sunday October 25, 2015. The Class Notes belongs to MATH 6118 at University of North Carolina - Charlotte taught by Staff in Fall. Since its upload, it has received 41 views. For similar materials see /class/228923/math-6118-university-of-north-carolina-charlotte in Mathematics (M) at University of North Carolina - Charlotte.
Reviews for Non
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/25/15
Chapter 2 The Rules of the Game We have mentioned that Euclid gathered all that was known about geometry and gathered it together in his Elements We note that from what we have found to date the Greeks beginning with Thales of Miletus took a more theoretical View of geometry and worked to systematize geometric knowledge by showing that certain results follow logically from others 21 Euclid s Elements We believe that the work of the Greek geometers reached a peak with the appearance of Euclid s work It appeared in 13 books where each book is more like one of a modern textbook s chapters Euclidean geometry is studied from three different perspectives that show themselves in high school geometry The rst and oldest of these is called the traditional perspective or sometimes the axiomatic perspective In this way of studying geometry we start as did Euclid with the axioms and postulates of Euclid From this core of results we can modify certain axioms or postulates and derive nite geometries non Euclidean geometries and other systems whose results can be compared to those that we deduced in Euclidean geometry A second perspective is credited to the German mathematician Felix Klein 184971925 who was looking for a manner in which to unify the study of geometry He together with the Swedish mathematician Sophus Lie 184271899 viewed geometry as studying properties of gures that are invariant under certain transformations This was not really new in some sense since Euclid used the property of superposition in his books This is called the transformation perspective of geometry By modifying the properties selected to be invariant we can generate different geometries 7 such as af ne geometries Euclidean geometry non Euclidean geometries projective geometries or even topology The third perspective also originates in the 19th century and is based on studying geometry as a vector space This is called the vector perspective It lends itself to a much more analytic study of geometry That being said any of these perspectives can be approached either synthetically or analytically The synthetic approaches tend to use numbers as rarely as possible and are based on a more axiomatic development of the geometry The analytic approaches take advantage of the properties of numbers and algebra to deduce geometric properties 19 20 CHAPTER 2 THE RULES OF THE GAME 22 Deduction and Proof The process by which we develop Geometry is tried and true We may not always start with exactly the same set of axioms or unde ned terms but we do know the path by which we will travel We will generate theorems using the rules of deductive logic77 What does this mean In logic especially the type that we will use in Geometry we accept the following common notion LAW OF THE EXCLUDED MIDDLE A statement is either true or false that is P or not P There is not other possibility For example if P is the proposition The grass is green then the Law of Excluded Middle holds that the logical statement Either Socrates is mortal or Socrates is not mortal is true by virtue of its form alone Here is n example of an argument that depends on the Law of Excluded Middle Lemma 21 There exist two irrational numbers a and I such that al7 is rational PROOF It is known that is irrational Consider the number 2 Clearly this number is either rational or irrational If it is rational we are done If it is irrational then let a and b Then ab n mm vs 2 and 2 is certainly rational This concludes the proof I In the above argument the statement this number is either rational or irrational invokes the Law of Excluded Middle There are multivalued logics or fuzzy logics in which we do not take the stance that each statement must be true or false This stance that you can always determine the truth of a mathematical statement is true in some sense If you agree with the assumptions in the statement or in the problem and if valid reasoning has been used the you must agree with the results This property of truth based on assumptions and valid reasoning is a result of a basic underlying aspect of mathematical thinking 7 the deductiue process or deduction ln mathematics results that have been deduced from agreed upon statements such as axioms de nitions or previously proven statements using valid arguments of deduction can be thought of as true We call these results theorems The deductive argument itself is called a proof Now before we messed it all up by nding non Euclidean geometries theorems proved using deduction were usually considered to be absolute truths or independent of the particular axioms that you chose for the geometry We sometimes create different axioms sets because some results may be more accessible using this axiom set rather than another We had thought that we always ended in the same place so there was little concern Now we know that the postulates that we chose de ne the mathematical system so our results are relative truths or true based on the assumptions of that system MATH 6118 090 Spring 2008 23 THE POWER OF DEDUCTION 21 23 The Power of Deduction It may seem hard to believe that this is such a powerful tool However it clearly trumps observation In Euclidean geometry we prove that the sum of the angles of a triangle is 180 This isn t just the triangles you can draw on your paper or the triangles that appear on the screen in a Dynamic Geometry System DGS or triangles that have whole number angle measures It is EVERY triangle You cannot nd every triangle and measure it 7 and even if you could you still could not measure it precisely enough We know that every triangle in Euclidean geometry has the same angle sum 7 every one There are times when we use deduction to prove the obvious77 such as that the base angles of an isosceles triangle are congruent However we must remember that what may seem obvious in one case may not be so obvious in the next Also sometimes what seems obvious may turn out not to be true Another strength is that by deduction we can prove theorems to be true even when they are hard to believe Another aspect of deduction is that it not only can show a statement to be true but also can indicate why it is true A fourth aspect of the power of deduction is that it provides a universally accepted criterion for the establishment of mathematical truth Although some philosophers may debate about the foundations of mathematics and some mathematicians question whether a proof in which computers did much of the work such as the solution to the Four Color Theorem constitutes a proof or whether a complicated proof has a gap there is universal agreement on the principles behind deductive proof Consequently when a new theorem is proved consider the proof of Fermat s Last Theorem or more recently the solution of the Poincare conjecture no one goes to the lab to check the results Only the argument is checked to see that the steps were valid based on the suppositions 24 All Power Corrupts With power comes responsibility It must be used carefully This is not always the case Geometry is especially suspect because we so much want to rely on drawings to see what is going on These can often lead one astrayll Theorem 21 Every triangle is isosceles PROOF Given AABC with AC 7 BC H Construct the bisector of C Call it t l is not perpendicular to E because AC 7 to Construct the perpendicular bisector of segment AB Call it m and let D AB O m Since 6 and m are not both perpendicular to AB 6 and m must intersect in a point OJ Construct the perpendicular from P to BC ancuet F be the foot of P in BC ewise construct the perpendicular from P to AC and let E be the foot of P in AC 4 ACP g BOP since 73 l is the bisector of lt0 5 CEP E CFP since they are both right angles MATH 6118 090 Spring 2008 MATH 6118 090 CHAPTER 2 THE RULES OF THE GAME 6 ACEP E ACFP by AAS 7 EP E FF and CE E OF because CPCTC l So7 AP E BP since a point on the perpendicular bisector of a line segment is equidis tant from the endpoints of the line segment 9 AAEP E ABFP since they are both right angles H 0 CLAIM EA E FB a Suppose that EA gt BF7 then there is a point A on EA7 different from A7 with EA E BF Then AA EP is a right angle This is impossible7 so EA E FB 11 AC E BC since CE E CF and EA U AC CE and EB U BC CF Therefore AABC is isosceles I Okay7 what is wrong If nothing is wrong7 then the result has to be true7 yet you can probably easily think of a triangle that is not isosceles Did we use an incorrect result No We any steps logically invalid No Yet something is wrong Here is another conundrum 7 you need to nd the mistake Theorem 22 A right angle has the same measure as an obtuse angle Spring 2008 ALL POWER CORRUPTS PROOF Construct a rectangle DABCD Choose a point E not on the rectangle so that AD E CD Construct the perpendicular bisector of AE and call it 6 Construct the perpendicular bisector of CD and call it m Then 6 O m 7 0 Call the point of intersection P Let M be the midpoint of AE and let N be the midpoint of CD Construct segments DP EP AP and CP H 1 H 1 2 03 BE 01 a T 1 2 AP E EP since P is on the perpendicular bisector of AE DP CP since P is on the perpendicular bisector of CD By construction AD E CE Thus AECP E AADP by the SSS Congruence criterion Therefore LECP E AADP by CPCTC Since DP E CP ADNP E ACNP by the SSS Congruence criterion ADCP E ACDP by CPCTC Now AECP ADCP AECD and AADP ACDP ZADC Thus AECD E AADC Since E lies outside the rectangle LECD gt ABCD so it is obtuse AACD is an angle of the rectangle so it is a right angle We have LECD E AADC so a right angle is congruent to an obtuse angle 1corresponding parts of congruent triangles are congruent MATH 6118 090 Spring 2008 24 CHAPTER 2 THE RULES OF THE GAME This completes the proof I Theorem 23 Two distinct pemendiculars can be drawn to a given line from a given ex temal point 5 U 21 PROOF Draw any two circles centered at O and O7 intersecting at points P and N Draw diameters PA and PB and then draw AB intersecting the circles at C and D APDA and APCB are right angles because they are inscribed in semicircles Thus PC and PD are both perpendicular to AB Theorem 24 Every point inside a circle is on the circle 1 PROOF Cgsider the circle 39y with center O and let P be inside the circle Choose a point B on ray OP so that OPOR 72 where r is the radius of 39y Let the perpendicular bisector of PR intersect the circle at points S and T and let M be the midpoint of PR OP OM 7 MP OROMMROMMP OPOR OM 7 MPOM MP 0M2 MP2 MATH 6118 090 Spring 2008 24 ALL POWER CORRUPTS 25 By the Pythagorean Theorem 0M2 MS2 7 OS2 OM 7 OS 7 MS MP2 MS PS2 MP 7 PS 7 MS 0P01 01 42 7 MP2 7 OS2 7 MSW 7 PS2 7 MSW 7 OS2 7 PS2 lt0Pgtlt0Rgt lt0Pgtlt0Rgt 7 P5 Therefore7 PS 0 and P must be on the circle I The purpose of this section is to show us that while the power of deduction is a very powerful tool7 we can misuse it by relying too much on diagrams This means that we must question many of the diagrams in proofs We should have an axiomatic or logical reason for a diagram to be represented in a proof It can be a guideline7 but it cannot serve as a proof itself MATH 6118 090 Spring 2008 Chapter 1 The Origins of Geometry 11 Introduction In the beginning geometry was a collection of rules for computing lengths areas and vol umes Many were crude approximations derived by trial and error This body of knowl edge developed and used in construction navigation and surveying by the Babylonians and Egyptians was passed to the Greeks The Greek historian Herodotus 5th century BC credits the Egyptians with having originated the subject but there is much evidence that the Babylonians the Hindu civilization and the Chinese knew much of what was passed along to the Egyptians The Babylonians of 2000 to 1600 BC knew much about navigation and astronomy which required a knowledge of geometry Clay tablets from the Sumerian 2100 BC and the Babylonian cultures 1600 BC include tables for computing products reciprocals squares square roots and other mathematical functions useful in nancial calculations Babyloni ans were able to compute areas of rectangles right and isosceles triangles trapezoids and circles They computed the area of a circle as the square of the circumference divided by twelve The Babylonians were also responsible for dividing the circumference of a circle into 360 equal parts They also used the Pythagorean Theorem long before Pythagoras per formed calculations involving ratio and proportion and studies the relationships between the elements of various triangles 12 The value of 7T The Babylonians also considered the circumference of the circle to be three times the diam eter Of course this would make 7139 3 7 a small problem This value for 7139 carried along to later times The Roman architect Vitruvius took 7139 3 Prior to this it seems that the Chinese mathematicians had taken the same value for 7139 This value for 7139 was sancti ed by the ancient Jewish civilization and sanctioned in the scriptures In 1 Kings 723 we nd He then made the sea of cast metal it was round in shape the diameter rim to rim being ten cubits it stood ve cubits high and it took a line thirty cubits long to go round it 7 The New English Bible 2 CHAPTER 1 THE ORIGINS OF GEOMETRY Rabbi Nehemiah attempted to change the value of 7139 to 227 but was rejected By 1800 BC the Egyptians according to the Rhind papyrus had the approximation 7139 z x 31604 To show how far we have progressed in the 1920 s the Indiana legislature passed a law mandating that 7139 It has since been appealed In 1789 Johann Lambert proved that 7139 is an irrational number and in 1882 F Lindemann proved that 7139 is transcendental ie it is not the solution to any algebraic equation with rational coefficients 121 Indiana and 7139 There are many false claims about the value of 7139 most of them coming from attempts to square the circle 7 one of the great unsolved problems of Greek mathematics One of the best stories is that of the Indiana legislature in 1897 The author of the bill was Edwin J Goodwin an MD of Solitude Indiana It seems that he was a crank or amateur mathematician who had been working on a procedure to square the circle He contacted his Representative one Taylor I Record of Posey County on January 18 1897 with his epoch making suggestion if the State would pass an Act recognizing his discovery he would allow all Indiana textbooks to use it without paying him a royalty What a nouel idea How could the legislature pass up such an o erf2 Nobody in the Indiana Legislature knew enough mathematics to know that the discouery was nonsense as it had been proved in the 1700 s that it is impossible to square the circle In due course the bill had its third House reading and passed 67 0 At this point the text of the bill was published and of course became the target for ridicule in this and other states The next day the following article appeared in the Indianapolis Sentinel To SQUARE THE CIRCLE Claims Made That This Old Problem Has Been Solved 77The bill telling how to square a circle introduced in the House by Mr Record is not intended to be a hoax Mr Record knows nothing of the bill with the exception that he introduced it by request of DrEdwin Goodwin of Posey County who is the author of the demonstration The latter and State Superintendent of Public Instruction Geeting believe that it is the long sought solution of the problem and they are seeking to have it adopted by the legislature Dr Goodwin the author is a mathematician of note He has it copyrighted and his proposition is that if the legislature will endorse the solution he will allow the state to use the demonstration in its textbooks free of charge The author is lobbying for the bill77 On 7 February 2 1897 Representative SE Nicholson of Howard County chairman of the Committee on Education reported to the House 77Your Committee on Education to which was referred House Bill No 246 entitled a a bill for an act entitled an act introducing a new mathematical truth has had same under consideration and begs leave to report the same back to the House with the recommendation that said bill do pass 7 The report was concurred in and on February 81897 it was brought up for the second reading following which it was considered engrossed Then 7Mr Nicholson moved that the constitutional rule requiring bills to be read on three days be suspended that the bill may be read a third time now7 The constitutional rule was suspended by a vote of 72 to 0 and the bill was then read a third time It was passed by a vote of 67 to O and the Clerk of the MATH 6118 090 Spring 2004 12 THE VALUE OF 7r 3 House was directed to inform the Senate of the passage of the bill The newspapers reported the suspension of the constitutional rules and the unanimous passage of the bill matter of factly except for one line in the Indianapolis Journal to the effect that this is the strangest bill that has ever passed an Indiana Assembly The bill was referred to the Senate on Feb 101897 and was read for the rst time on Feb 11 and referred to the Committee on Temperance On Feb12 Senator Harry S New of Marion County Chairman of the Committee on Temperance made the following report to the Senate Your committee on Temperance to which was referred House Bill No 246 introduced by MrRecord has had the same under consideration and begs leave to report the same back to the Senate with the recommendation that said bill do pass The Senate Journal mentions only that the bill was read a second time on Feb 12 1897 that there was an unsuccessful attempt to amend the bill by strike out the enacting clause and nally it was postponed inde nitely That the bill was killed appears to be a matter of dumb luck rather than the superior education or wisdom of the Senate It is true that the bill was widely ridiculed in Indiana and other states but what actually brought about the defeat of the bill is recorded by Professor C A Waldo in an article he wrote for the Proceedings of the Indiana Academy of Science in 1916 The reason he knows is that he happened to be at the State Capitol lobbying for the appropriation of the Indiana Academy of Science on the day the Housed passed House Bill 246 When he walked in the found the debate on House Bill 246 already in progressln his article he writes according to Edington An eX teacher from the eastern part of the state was saying 7The case is perfectly simple If we pass this bill which establishes a new and correct value for 7139 the author offers to our state without cost the use of his discovery and its free publication in our school text books while everyone else must pay him a royalty7 The roll was then called and the bill passed its third and nal reading in the lower house A member then showed the writer ie Waldo AA a copy of the bill just passed and asked him if he would like an introduction to the learned doctor its author He declined the courtesy with thanks remarking that he was acquainted with as many crazy people as he cared to know That evening the senators were properly coached and shortly thereafter as it came to its nal reading in the upper house they threw out with much merriment the epoch making discovery of the Wise Man from the Pocke Note that this value of 7139 is mentioned in the Guiness Book of Records 7 as the most inaccurate value for 7r It appears that Dr Goodwin was not satis ed with just one value of 7139 as the following values appear in the bill Since the rule in present use presumably 7r equals 314159 fails to work it should be discarded as wholly wanting and misleading in the practical applications the bill de clared lnstead mathematically inclined Hoosiers could take their pick among the following formulae H The ratio of the diameter of a circle to its circumference is 54 to 4 In other words 7r 165 32 to The area of a circle equals the area of a square whose side is 14 the circumference of the circle Working this out algebraically we see that 7139 must be equal to 4 MATH 6118 090 Spring 2004 4 CHAPTER 1 THE ORIGINS OF GEOMETRY 3 The ratio of the length of a 90 arc to the length of a segment connecting the arc s two endpoints is 8 to 7 This gives us 7139 xEm167 or about 323 There may have been other values for 7139 as well the bill was so confusingly written that it s impossible to tell exactly what Dr Goodwin was getting at Mathematician David Singmaster says he found six different values in the bill plus three more in Goodwin s other writings and comments for a total of nine 13 Egyptian Geometry Now the Egyptians were not nearly as inventive as the Babylonians but they were extensive users of mathematics especially geometry They were extremely accurate in their construc tion making the right angles in the Great Pyramid of Giza accurate to one part in 27000 From the above approximation they computed the area of a circle to be the square of 89 of the diameter 2 2 8 16 2 A 6 1 3 T They also knew the Pythagorean Theorem and were able to compute volumes and dihedral angles of pyramids and cylinders 14 Early Greek Geometry The ancient knowledge of geometry was passed on to the Greeks They seemed to be blessed with an inclination toward speculative thinking and the leisure to pursue this inclination They insisted that geometric statements be established by deductive reasoning rather than trial and error This began with Thales of Milete He was familiar with the computations right or wrong handed down from Egyptian and Babylonian mathematics In determining which ofthe computations were correct he developed the rst logical geometry This orderly development of theorems by proof was the distinctive characteristic of Greek mathematics and was new He is credited with proving the following results H A circle is bisected by any diameter to The base angles of an isosceles triangle are equal 03 The angles between two intersecting straight lines are equal q Two triangles are congruent if they have two angles and one side equal 01 An angle inscribed in a semicircle is a right angle This new mathematics of Thales was continued over the next two centuries by Pythago ras and his disciples The Pythagoreans as a religious sect believed that the elevation of the soul and union with God were achieved by the study of music and mathematics Nonetheless they developed a large body of mathematics by using the deductive method Their foundation of plane geometry was brought to a conclusion around 400 BC in the Ele ments by the mathematician Hippocrates This treatise has been lost but many historians agree that it probably covered most of Books l lV of Euclid s Elements which appeared about a century later circa 300 BC MATH 6118 090 Spring 2004 15 EUCLlDEAN GEOMETRY 5 Euclid was a disciple of the Platonic school Around 300 BC he produced the de nitiue treatment of Greek geometry and number theory in his thirteen uolume Elements In com piling this masterpiece Euclid built on the experience and achievements of his predecessors in preceding centuries on the Pythagoreans for Books I IV VII and IX on Archytas for Book VIII on Eudoxus for Books V VI and X11 and on Thecetetus for Books X and XIII So completely did Euclid s work supersede earlier attempts at presenting geometry that few traces remain of these e orts It s a pity that Euclid s heirs have not been able to collect royalties on his work for he is the most widely read author in the history of mankind His approach to geometry has dominated the teaching of the subject for ouer two thousand years Moreouer the axiomatic method used by Euclid is the prototype for all of what we now call pure mathematics It is pure in the sense ofpure thought no physical experiments need be performed to verify that the statements are correct only the reasoning in the demonstra tions need be checked In this treatise he organized a large body of known mathematics including discoveries of his own into the rst formal system of mathematics This formalness was exhibited by the fact that the Elements began with an explicit statement of assumptions called axioms or postulates together with de nitions The other statements 7 theorems lemmae corollaries 7 were then shown to follow logically from these axioms and de nitions Books l lV VII and IX of the work dealt primarily with mathematics which we now classify as geometry and the entire structure is what we now call Euclidean geometry 15 Euclidean Geometry Euclidean geometry was certainly conceived by its creators as an idealization of physical geometry The entities of the mathematical system are concepts suggested by or abstracted from physical experience but differing from physical entities as an idea of an object differs from the object itself However a remarkable correlation existed between the two systems The angle sum of a mathematical triangle was stated to be 180 if one measured the angles of a physical triangle the angle sum did indeed seem to be 180 and so it went for a multitude of other relations Because of this agreement between theory and practice it is not surprising that many writers came to think of Euclid s axioms as self evident truths Centuries later the philosopher Immanuel Kant even took the position that the human mind is essentially Euclidean and can only conceive of space in Euclidean terms Thus almost from its inception Euclidean geometry had something of the character of dogma Euclid based his geometry on ve fundamental assumptions Postulate I For euery point P and for euery point Q not equal to P there exists a unique line 6 that passes through P and Postulate II For euery segment AB and for euery segment CD there exists a unique point E such that B is between A and E and segment CD is congruent to segment BE MATH 6118 090 Spring 2004 6 CHAPTER 1 THE ORIGINS OF GEOMETRY Postulate III For every point 0 and every point A not equal to 0 there exists a circle with center 0 and radius OA Postulate IV All right angles are congruent to each other Before we study the Fifth Postulate let me say a few words about his de nitions Euclid s methods are imperfect by modern standards He attempted to de ne everything in terms of a more familiar notion sometimes creating more confusion than he removed As an example A point is that which has no part A line is breadthless length A straight line is a line which lies evenly with the points on itself A plane angle is the inclination to one another of two lines which meet When a straight line set upon a straight line makes adjacent angles equal to one another each of the equal angles is a right angle Euclid did not de ne length distance inclination or set upon Once having made the above de nitions Euclid never used them He used instead the rules of interaction between the de ned objects as set forth in his ve postulates and other postulates that he implicitly assumed but did not state Postulate V 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 inde nitely meet on that side on which the angles are less than two right angles No one seemed to like this Fifth Postulate possibly not even Euclid himself he did not use it until Proposition 29 The reason that this statement seems out of place is that the rst four postulates seem to follow from experienceitry to draw more than one line through 2 different points The Fifth Postulate is unintuitive It does come from the study of parallel lines though Equivalent to this postulate is Playfair397s Postulate Given a line and a point not on that line there exists one and only one line through that point parallel to the given line Euclid s Fifth Postulate seemed to be too burdensome It should follow from the other axioms Since it is not intuitive we should be able to prove it as a theorem We should be able to prove that it is dependent in this Axiom system If we have a set of axioms A1 A2 An for our mathematical system and we can prove that Axiom An is derivable or provable from the other axioms then An is indeed redundant In some sense we are looking for a basis for this mathematical system Unlike vector spaces and linear algebra there is not a unique number of elements in this basis for it includes the axioms de nitions and the rules of logic that you use Many people have tried to prove the Fifth Postulate The rst known attempt to prove Euclid V as it became known was by Posidionius1st century BC He proposed to replaced the de nition of parallel lines those that do not intersect by de ning them as coplanar lines that are everywhere equidistant from one another It turns out that without Euclid V you cannot prove that such lines exist It is true that such a statement that parallel lines are equidistant from one another is equivalent to Euclid V Ptolemy followed with a proof that used the following assumption For every line 6 and every point P not on 5 there exists at most one line m through P such that m is parallel to l We will show at a later date that this statement is equivalent to Euclid V and therefore this did not constitute a proof of Euclid V MATH 6118 090 Spring 2004 15 EUCLlDEAN GEOMETRY 7 Proclus 410 485AD also attempted to prove Euclid V His argument used a limiting process He retained all of Euclid s de nitions all of his assumptions except Euclid V and hence all of his propositions which did not depend on Euclid V His plan was 1 to prove on this basis that a line which meets one of two parallels also meets the other and 2 to deduce Euclid V from this proposition His handling of step 2 was correctly handled The argument in step 1 runs substantially as follows Let g and h be parallel lines and let another line k meet h at P From Q a point of k situated between 9 and h drop a perpendicular to h As Q recedes inde nitely far from P its distance QR from h increases and exceeds any value however great In particular QR will exceed the distance between 9 and h For some position of Q then QR will equal the distance between 9 and h When this occurs k will meet 9 There are a number of assumptions here which go beyond those found in Euclid I will mention only the following two 0 the distance from one of two intersecting lines to the other increases beyond all bounds as we recede from their common point 0 the distance between two parallels never exceeds some nite value The rst of the two assumptions is not a grave error on the part of Proclus for it can be proved as a theorem on the basis of what he assumed from Euclid Unfortunately for Proclus his second assumption is equivalent to Euclid V Nasiraddin 1201 1274 John Wallis 1616 1703 Legendre 1752 1833 Wolfgang Bolyai Girolamo Saccheri 1667 1733 Johann Heinrich Lambert 1728 1777 and many others tried to prove Euclid V and failed In these failures there developed a goodly number of substitutes for Euclid V ie statements that were equivalent to the statement of Euclid V The following is a list of some of these that are more common H Through a point not on a giuen line there passes not more than one parallel to the line to Two lines that are parallel to the same line are parallel to each other A line that meets one of two parallels also meets the other m If two parallels are cut by a transuersal the alternate interior angles are equal 01 There exists a triangle whose angle sum is two right angles a Parallel lines are equidistant from one another There exist two parallel lines whose distance apart neuer exceeds some nite ualue 00 1 Similar triangles exist which are not congruent to Through any three non collinear points there passes a circle H 0 Through any point within any angle a line can be drawn which meets both sides of the angle 11 There exists a quadrilateral whose angle sum is four right angles 12 Any two parallel lines have a common perpendicular MATH 6118 090 Spring 2004 8 CHAPTER 1 THE ORIGINS OF GEOMETRY lt fell to three different mathematicians to independently show that Euclid V is not provable from the other axioms and what is derivable from them These were Carl Friedrich Gauss Jnos Bolyai and Nicolai lvanovich Lobachevsky Once these men broke the ice the pieces of geometry began to fall into place More was learned about non Euclidean geometriesihyperbolic and elliptic or doubly elliptic The elliptic geometry was studied by Riemann gave rise to riemannian geometry and manifolds which gave rise to differential geometry which gave rise to relativity theory et al This gives us something to anticipate as we learn more about geometry We will spend our time studying hyperbolic geometry for it lends itself to better studyinot requiring major changes in the axiom system that we have chosen We may have an opportunity to see that hyperbolic geometry is now lending itself to considerations in the latest research areas of mathematics MATH 6118 090 Spring 2004 Chapter 7 Other Geometries 71 The Idea of Parallelism We have agreed that we would work with a reasonable set of axioms for our geometry We require that our sets of axioms be consistent independent and complete De nition 71 A set of axioms is said to be consistent if neither the axioms nor the propo sitions of the system contradict one another De nition 72 A set of axioms is said to be independent if none of the axioms can be derived from any of the other axioms De nition 73 A set of axioms is said to be complete if it is not possible to add a new independent axiom to the system We have looked at Euclid s axioms and have commented on how the rst four differ from the Fifth Axiom in that they are direct concise and easy to read The efforts of mathematicians since Euclid s time to show that the Fifth Axiom is dependent on the rst four have all met with failure Geometer s have offered 77proofs77 of the Fifth Axiom but each has been found to be awed This is one of the times that failure has been helpful in that they all turn out to be equivalent statements to Euclid s Fifth Axiom We are able to show that each of the following statements is logically equivalent to Euclid s Fifth Axiom Playfair s Postulate Through a point not on a given line exactly one parallel may be drawn to the given line H The sum of the angles in a triangle is equal to two right angles to There exists a pair of similar triangles that are not congruent OJ There exists a pair of lines everywhere equidistant from one another F If three angles of a quadrilateral are right angles then the fourth angle is also a right angle 01 If a line intersects one of two parallel lines it will intersect the other 50 72 SACCHERI QUADRILATERALS 51 6 Lines parallel to the same line are parallel to one another 7 Two lines which intersect one another cannot both be parallel to the same line Now we would need to show that each of these is equivalent to Euclid s Fifth Axiom by showing that each implies that Euclid s Fifth Axiom holds and that Euclid s Fifth Axiom implies that each of these is true You are allowed to use any of the Propositions that Euclid proved without using his Fifth Axiom This is an exercise in itself and not where we want to spend our time One of the mathematicians who worked on proving that Euclidean geometry was the only geometry was the Italian priest and mathematician Giovanni Girolamo Saccheri 72 Saccheri Quadrilaterals Saccheri was a Jesuit priest and university professor living from 1667 to 1733 Before he died he published a book entitled Euclides ab omni nceuo uindicatus Euclid Freed of Euery Flaw lt sat unnoticed for over a century and a half until rediscovered by the Italian mathematician Beltrami He wished to prove Euclid s Fifth Postulate from the other axioms To do so he decided to use a reductio ad absurdum argument He assumed the negation of the Parallel Postulate and tried to arrive at a contradiction He studied a family of quadrilaterals that have come to be called Saccheri quadrilaterals Let S be a convex quadrilateral in which two adjacent angles are right angles The segment joining these two vertices is called the base The side opposite the base is the summit and the other two sides are called the sides If the sides are congruent to one another then this is called a Saccheri quadrilateral The angles containing the summit are called the summit angles We are able to prove the following Theorem 71 In a Saccheri quadrilateral i the summit angles are congruent and ii the line joining the midpoints of the base and the summiticalled the alt itudei is perpendicular to both A M 3 Now Saccheri studied the three different possibilities for these summit angles Hypothesis of the Acute Angle HAA The summit angles are acute Hypothesis of the Right Angle HRA The summit angles are right angles Hypothesis of the Obtuse Angle HOA The summit angles are obtuse Saccheri intended to show that the rst and last could not happen hence he would have found a proof for Euclid s Fifth Axiom He was able to show that the Hypothesis of the Obtuse Angle led to a contradiction of what is now know as the Saccheri Legendre Theorem MATH 6118 090 Spring 2004 52 CHAPTER 7 OTHER GEOMETRlES see below He was unable to arrive at a contradiction when he looked at the Hypothesis of the Acute Angle He gave up rather than accept that there was another geometry available to study Theorem 72 SaccheriLegendre Theorem The sum of the degree measures of the three angles in any triangle is less than or equal to 180 4A 4B 40 180 It took another 150 years or so for someone to nally accept the existence of other geometries Separately Janos Bolyai and Nicolai Lobachevsky discovered new geometries realized what they had done and published their results They approached the question of parallelism through Playfair s Postulate Again they had to consider three possibilities 7 and each one leads to a different geometry Hyperbolic Axiom Through a point not on a given line more than one parallel may be drawn to the given line Euclidean Axiom Through a point not on a given line exactly one parallel may be drawn to the given line Elliptic Axiom Through a point not on a given line no parallels may be drawn to the given line The geometry that they studied was the rst published logically consistent alternative to Euclid Their discovery is known as hyperbolic or Lobachevskian geometry and is characterized by the following axioms 0 Given any two points exactly one line may be drawn containing these points 0 Given any line a segment of any length may be determined on the line 0 Given any point a circle of any radius may be drawn 0 All right angles are congruent 0 Through a point not on a given line at least two parallel lines may be drawn to the given line 73 Poincar s Disk Model for Hyperbolic Geometry If we adopt the Hyperbolic Axiom then there are certain rami cations 1 The sum of the angles in a triangle is less than two right angles 2 All similar triangles that are congruent ie AAA is a congruence criterion 3 There are no lines everywhere equidistant from one another 4 If three angles of a quadrilateral are right angles then the fourth angle is less than a right angle 5 If a line intersects one of two parallel lines it may not intersect the other 6 Lines parallel to the same line need not be parallel to one another MATH 6118 090 Spring 2004 73 POlNCARE S DISK MODEL FOR HYPERBOLIC GEOMETRY 53 7 Two lines which intersect one another may both be parallel to the same line How can we see this It cannot be by just looking at the Euclidean plane in a slightly different way We would like a model with which we could study the hyperbolic plane If it is to be a Euclidean object that we use to study the hyperbolic plane H2 then we must have to make some major changes in our concept of point line and distance We need a model to see what H2 looks like We know that it will not be too easy but we do not want some extremely difficult model to construct We will work with a small subset of the plane but give it a different way of measuring distance In order to give a model for H2 we need to determine the set of points then determine what lines are and how to measure distance For Poincare s Disk Model we take the set of points that lie inside the unit circle ie the set H2 mv l2v2lt1l Note that points on the circle itself are NOT in the hyperbolic plane However they do play an important part in determining our model Euclidean points on the circle itself are called ideal points omega points vanishing points or points at in nity A unit circle is any circle in E21 is a circle with radius one De nition 74 Given a unit circle 2 in the Euclidean plane points of the hyperbolic plane are the points in the interior ofE Points on this unit circle are called omega points of the hyperbolic plane If we take 2 to be the unit circle centered at the origin then we would think of the hyperbolic plane as H2 l x2 y2 lt 1 and the omega points are the points 9 l z2y2 1 The points in the Euclidean plane satisfying x y l x2y2 gt 1 are called vltraideal points We now have what our points will be We see that we are going to have to modify our concept of line in order to have the Hyperbolic Axiom to hold De nition 75 Given a unit circle 2 in the Euclidean plane lines of the hyperbolic plane are arcs of circles drawn orthogonaF to E and located in the interior of 2 731 Construction of Lines This sounds nice but how do you draw them H Start with a circle P centered at O and consider the ray to Construct the line perpendicular to Q at A 03 Choose a point P on this perpendicular line for the center of the second circle and make PA the radius of a circle centered at P F Label the second point of intersection with circle P B Then the arc AB represents a line in this model 1the Euclidean plane 2Circles are orthogonal to one another when their radii at the points of intersection are perpendicular MATH 6118 090 Spring 2004 54 CHAPTER 7 OTHER CEOMETRIES Figure 72 Poincare lines through A Now7 how do you construct these lines in different cir cumstances There are three cases we need to consider Case I A7 B E P Case II A E P and B lies inside P Case III A and B both lie inside P Case I Construct rays 17A and P B where P is the cermir of thgircle P Construct the lines perpendicular to PA and PB at A and B respectively Let Q be the point of intersection of those two lines The circle 9 centered at Q with radius QA intersects P at A and B The line between A and B is the arc of 9 that lies inside P Note that this arc is clearly orthogonal to P by its con struction Case II Construct rays 17A and P B where P is the center of the circle P Construct the line perpendicular to 17A at A Draw segment AB and construct its perpendicular bisector Let Q be the point of intersection of this line and the tangent line to P at A The circle 9 centered at Q with radius QA contains A and B The line containing A and B is the arc of 9 that lies inside P This arc7 as constructed is orthogonal to P at A We want to see that it is orthogonal at the other point of intersection with the circle Let that point of intersection be X Then7 X E P means that PA E PX Since X lies on our second circle it follows that QX E QA Since PQ E PQ7 we have that APAQ E APXQ7 which means that APXQ is a right angle7 as we wanted to show Case III Construct the ray 17A and then construct the line perpendicular to 17A at A This intersects P in points X and Y Construct the tangents to P at X and at Y These tangent lines intersect at a point C The circle 9 centered at Q is the circle passing through A7 B7 and C The line containing A and B is the arc of 9 that lies inside P Figure 71 Poincare line MATH 6118 090 Spring 2004 73 POINCARE S DISK MODEL FOR HYPERBOLIC GEOMETRY 55 From our construction7 we have that APXC N APAX and it follows that lPAHPCl lPXlZ r2 Now7 Q lies on the perpendicular bisectors of AC and AB as Q is the circumcircle for AABC There is a point T on the circle 9 so that the tan gent line to Q at T passes through P Construct the line through P and Q which intersects the circle in two points G1 and G2 so that G1 lies between P and Q Now 7 lPTlZ lPQlZ lQTlZ Figure 73 Poincare line in Case 111 lPQl lQTl lPQl lQTl lPQl lQGIl lPQl lQGzl lPGlllPGgl which by Theorem 537 lPAHPCl r2 Therefore7 T lies on the circle P and P and Q are orthogonal at that point A similar argument shows that they are orthogonal at the other point of intersection 732 Distance Now7 this Euclidean area inside the unit circle must represent the in nite hyperbolic plane This means that our standard distance formula will not work We introduce the distance metric by 2dr 1 7 72 where p represents the hyperbolic distance and r is the Euclidean distance from the center of the circle Note that dp a 00 as r a 1 This means that lines are going to have in nite extent The relationship between the Euclidean distance of a point from the center of the circle and the hyperbolic distance is T Qdu 17 1 p017u2lnlt gt2tanh r Now7 we can use this to de ne the distance between two points on a Poincare line Given two hyperbolic points A and B7 let the Poincare line intersect the circle in the omega points P and Q Let dp or r tanhE 7 APAQ 7 AP BQ T BPBQ 7 AQ BP7 denote the cross ratio of A and B with respect to P and Q7 where AP denotes the the Euclidean arclength De ne the hyperbolic distance from A to B to be AB7 PQ dA B log lAB PQl MATH 6118 090 Spring 2004 56 CHAPTER 7 OTHER GEOMETRIES Theorem 73 If a point A in the interior of E is located at a Euclidean distance r lt 1 from the center 0 its hyperbolic distance from the center is given by 1 r d A O l 7 7 0g 17 T Theorem 74 The hyperbolic distance from any point in the interior of E to the circle itself is in nite 733 Parallel Lines gt It is easy twe that the Hyperbolic Axiom works in this model Given a line AB and a point D AB then we can draw at least two lines through D that do not intersect AB Figure 74 Multiple parallels through A Call these two lines through D lines 1 and 2llotice now how two of our results do not hold as we remarked earlier We have that AB and 1 and AB and 2 are parallel but 1 and 2 are not parallel Note also that 2 int ersects one of a pair of parallel lines 61 but does not intersect the other parallel line AB The hyperbolic plane has two types of parallel lines The de nition that we will give here will depend explicitly on the model that we havmosen but we will make a more general de nition later Consider the hyperbol ine AB which intersects the circle 2 in the ideal points A and 9 Take a point D AB Construct the line through A and D Since this line does not intersect the line AB inside the circle these two hyperbolic lines are parallel However they seem to be approaching one another as we go 77to in nity ln somegense this is true as we shall see later nce there are two 77ends77 of the Poincare line AB there are two of these lines The line AB and DA are said to be limiting parallel3 The de ning property is as follows De nition 76 Let P 6 AB Consider the collection of lines as P goes to Q or A The rst line thrgugh D in this collection that does not intersect AB in H2 is the limiting parallel line to AB in that direction 3These are also called sensedeparallel ultraparallel or horoparallel MATH 6118 090 Spring 2004 73 POlNCARFS DISK MODEL FOR HYPERBOLIC GEOMETRY 57 Drop a perpendicular from D to APB and label this point of intersection M Angles AADM and AQDM are called angles of parallelism Theorem 75 The angles of parallelism associated with a given line and point are congruent PROOF Assume not ie assume AADM 7 AQDM Then one angle is greater than the other Without loss of generality we may assume that AADM lt AQDM Then there is a point E in the interior of AQDM such that AADM AEDM The line ED must intersect AB since DD is the limiting parallel line to AB in that direction Let the point of intersection be F Choose G on AB on the opposite side of DM from F so that Pomcare Llnes FM GM Then AGMD E AFMD This implies that AGDM AFDM AADM This means that DQ in ter A sects AB at G This contradicts the condition that DD is limiting parallel to AB Thus the angles of parallelism are congruent I Figure 75 Limiting Parallel Theorem 76 The angles of parallelism associated with a given line and point are acute PROOF Assume not ie assume that AMDQAgt 90 ien there is a point E interior to AMDQ so that AMDE 90 Then since DE and AB are perpendicular to the same line they are parallel Thus DE does not intersect AB which contradicts the condition that DD is the limiting parallel line If the angle of parallelism is 90quot then we can show that we have Euclidean geometry Thus in H2 the angle of parallelism is acute Theorem 77 Lobaclyevskifs Theorem Given a point P at a hyperbolic distance d from a hyperbolic line AB ie dP M d the angle of parallelism 0 associated with the line and the point satis es ifd tan 2 lim0 and lim 00 daO 2 daoo Note then that PROOF The proof of this is interesting in that we play one geometry off of the other in order to arrive at our corglusion We are given a line ABAand a point P not on the line Construct the line through P which is perpendicular to AB Call the point of intersection R as in Figure 76 Then we have that d dP R We can translate P to the center of the unit circle and translate our line to a line so that our line perpendicular to AB is a radius of E as we have done in Figure 77 Construct the radii from P to the ideal points A and B and construct the line s tangent to E at these points These tangent lines intersect at a point Q which lies on PR Now since we have moved our problem to the center of the circle we can use our previous result to see that if r is the Euclidean distance from P to R then we have 17 ollog1 77 MATH 6118 090 Spring 2004 58 CHAPTER 7 OTHER GEOMETRIES Figure 76 Figure 77 or rewriting this we have ed1r ore d1ir 1 7 r 1 l r Now7 we are talking about Euclidean distances with r and using our Euclidean right triangles with radius 1 we have that T QILQR QPi QASBCZQPAitanZQPAseceitane F781quot COS Now7 algebra leads us to 7d 177 17 cos0sin071 cos07sin01 7 cos0sin071cos0sin01 T cos07sin01cos0sin01 cos20 2cos0sin0sin2071 cos20 2cos07sin201 2sin0cos0 sin0 2cos20 2cos0 1cos0 2sin cos 734 Hyperbolic Circles Now7 if we have a center of a circle that is not at the center P of the unit circle 2 we know that the hyperbolic distance in one direction looks skewed with respect to the Euclidean distance That would lead us to expect that a circle in this model might take on an elliptic or oval shape We will prove later that this is not the case In fact7 hyperbolic circles embedded in Euclidean space retain their circular appearance 7 their centers are offset MATH 6118 090 Spring 2004 73 POINCARE S DISK MODEL FOR HYPERBOLIC GEOMETRY 59 Theorem 78 Giuen a hyperbolic circle with radius R the circumference C of the circle is giuen by C 27139 sinhR 735 Similarities With Euclidean Geometry Because we only changed the Fifth Axiom and not the rst four7 everything that holds in Neutral Geometry or geometry without a parallel postulate also holds in the Hyperbolic Plane Other things may also hold7 but may require a different proofll Examples oftheorems that are still true are Theorem 79 Pasch7s Theorem If AABC is any triangle and l is any line intersect ing side AB in a point between A and B then 6 also intersects either side AC or side BC If C 6 then 6 does not intersect both AC and BC A A A A Theorem 710 Crossbar Theorem IfAD is between AB and AC then AD intersects the segment BC Theorem 711 Supplements of congruent angles are congruent Theorem 712 Vertical angles are congruent to each other Theorem 713 An angle congruent to a right angle is a right angle Theorem 714 For euery line 6 and euery point P there exists a line through P perpen dicular to 5 Theorem 715 ASA Giuen AABC and ADEF with AA E 1D 1C E 1F and AC E DF Then AABC E ADEF Theorem 716 SSS Giuen triangles AABC and ADEF If AB E DE AC E DF and BC E EF then AABC E ADEF Theorem 717 Alternate Interior Angles Theorem If two lines cut by a transuer sal have a pair of congruent alternate interior angles then the two lines are non intersecting Theorem 718 Ifm and n are distinct lines both perpendicular to the line 6 then m and n are non intersecting Theorem 719 If P is a point not on t then the perpendicular dropped from P to t is unique Theorem 720 If t is any line and P is any point not on 6 there exists at least one line m through P which does not intersect 6 Theorem 721 Exterior Angle Theorem An exterior angle of a triangle is greater than either remote interior angle Theorem 722 SAA Congruence In triangles AABC and ADEF given that AC E DF 4A a 4D and 4B 4E then AABC ADEF MATH 6118 090 Spring 2004 60 CHAPTER 7 OTHER GEOMETRIES Theorem 723 Two right triangles are congruent if the hypotenuse and a leg of one are congruent respectively to the hypotenuse and a leg of the other Theorem 724 Euery segment has a unique midpoint Theorem 725 Euery angle has a unique bisector Theorem 726 Euery segment has a unique perpendicular bisector Theorem 727 In a triangle the greater angle lies opposite the greater side and the greater side lies opposite the greater angle ie AB gt BC if and only if AC gt 1A Theorem 728 Giuen AABC and AABC if AB AB and BC B C then 1B lt ZB if and only if AC lt A C MATH 6118 090 Spring 2004 MATH 6118 Collinzar ify 252008 There are Three kinds of Those who can counf and Those who can39f A Circumcen rer a A 252008 Cemmid Incenter L 1 The 4Cemers so far 252008 The Euler Segment luvmnm u quotmm m htuhmmmmu lrmv zz 06 The Euler Segme 1 mm rmmumka Emnnnmum mm my mm 2011 mmmmmmmw Mum n mm um quotmm mm 2r 20mm 2 mam mm n m MEN 2 Am M mama mummy imammmaw mu m r Mamawwwmmmmmm mum111mm mm am 2 The Pedal Triangle v L z mummnmunmmgmnmum mmlamnnnm mmupummnmmm mmmwmmmmw The Pedc Trmng e fer The ercumcw 252008 252008 The SW50quot Lme Th mm Thefee of The perpenmcmqrs fromapmmmhw sofamangTeare cohmear w and my w The pmm he on The mrcumcwde The SW50quot Lme Fm assume he cwmmcwc e WLOEweccn assume mm P TS m we mm dues rm mmam Band P TS an em cs for m cs w TsfrumA hacessar uccn rewdaehmmmsm I make m5 5 H3 The SW50quot Lme thsnmm mumunhnhmmg z AYEszme mew gxs mmmmu LthLqummh39zmThy n ussz human men g P Lmszv has an m mumm n1 AYZC am My 252008 The Swmsan Lme wumuw 4va The Swmsan Lme 2s 2 my Mung We pmms caHmBar The ergannern r m EFb2m2 pa 2mm A n Ag Tmth m3 BE an F 252008 The ergannern r AFAE E A BFBD n D E beamsuheyare A A Bx ermhmgems By am Theorem Meyare concurrem The LemmnePamT re ecmns of L memoquot across m2 assammed ang e manna1 b SB or I The LemmnePamT WWW Md 5 mum m n Mum hwtmnm nm 5 I my wmx mrkwu r wnwwnnmd hw mmwa x in 2 a l mumm 252008 The LemmherhT WWW as md 55 Mum m a pm mm m Lemur1 pm m Wewl mrkwuxw mmmmmw mmqu Lima 2 a l mumm The LemmherhT The LemmherhT x whp y hwxw yawn md mgmmm w gt mmquot mum c m 5qu n5 27H kgwnAdngAQkH mm Wm M mm d u 3 Xaqv My x y Wm Wm M mum Mums a sum 4mm gt 5mm mmmaq w MM Wm M W x Lmodya nNaqi A x wwudamaq x w y Lmot a nNaqi a m WmnuusxPAW WWW WNW Z W m pkgwuwu wwwnw zavvwa wxodwwad 2x41 SUUZSZ Mu WWW Aawmu 2 mm pmNam w mm 3 Mudwmdga z mu m wtdsmud midsaw mm umwm aw mum r w M m Wm M x wam ag a nN mi mvvwwud mm m Wm 49 5mm m mud quotM mm m am am a Lmodya nNaqi SUUZSZ mm mum an 7 mam m 77 Aux mm m7 m2 waxwmwmguxn m2 mm7wm maqoaqis anbxw uxx am am 7 am u an mm cam 27 am u a w www 4 Wm mum 2 w Whammy mum mame mud human 04 WW a wuum amvv m maqoaqis anbxw 0 mm pmwm Mp7 u Wm WM upquot WWW Axanwdwav M W 721 mm M 0 an maqoaqis anbxw SUUZSZ mum m m ma w mm m uuuuwum y muumw uum mg a 39w m m u in w x mp HHHV may My wwwpmmw mamaan mug WWW pm 0 dawnxxy mam slsnnyauaw E E W mm m M quotWm 721W y WWW 8 M 21v ivxanxwmapw o ammwm mam slsnnyauaw mm Mum u miname a mum mm unawqu y man awn quuxuuumlpumu mamqi slmqow SUUZSZ a a WWI mm mm mm mum x mm mm mm m 721W kamxuamu u pm MquM 0pm mm mm awnxw wanmaaobv wwmwud ammuanst Mum WW maman mam slsnnyauaw SUUZSZ MATH 6118 Rules of The Game 252008 Everyone is a fool for at least five minuTes a clay WISDOM consisTs or noT exceeding The limit The Power of Deduc iion Theorem There exisi iwo irraiiohai numbers a and b so ihai ah is raiiohai We know ihai f2 is irraiiohai Consider m number a 2 A is eiiher raiiohai 07 irraiioriai If ii is raiiohai WE are Vi SO assume ihai ii is irraiioriai Lei b J2 Then 5 2 2J3 2 ab Ja 5 5 2 which is cieariy raiiohai w u 21v q vwww Wwavw umumwmm mm 3 WWW mum sawan sx ay uwi W3 gymmwmd swiuunua mu umnqamwwu x mwwumavvuaw mm 3 WWW mum Mawnsqv mm mm WW Mmmaw paau w s am m sum mm W aoozsz mveam 5 WM 4 wuwmumm i mpmmum v auwmwm 2 wam Wm M w mu umumwmm Mummy 2 gymmwmd muunu mu auwmwm 2 wuavq www Wwaawu umqmwmd Mummy 2 gymmwmd mwnu mu mmmmw x mwwumavvuaw mm 3 WWW mum avum s wuwmumm i mpmmum v auwmwm 2 wuam Wm M w mu umumwmm Mummy 2 gymmwmd muunu mu sawan sx ay uwi W3 aoozsz mam igggjwudj 2 WM um Luv 721 um um Aumwadnduduq v 1 nu w mwm 2 sawan sx ay uwi W3 SW MMWIMMM m 4477 77 v avum um Luv 721 um um Aumwadnduduq v auwmwm 2 wam Wm M w mu umumwmm Mummy 2 sawan sx ay uwi W3 aoozsz A myme WWW M 4 v m m m 21 n mm m m w mmuq mmwm WW7 yaw um Luv 721 um um Aumwadnduduq v 1 nu w mwm 2 sawan sx ay uwi W3 A m 1 mm mm ma 7 w a avum um Luv 721 um um Aumwadnduduq v 1 nu w mwm 2 sawan sx ay uwi W3 aoozsz 252003 Every mnng s Isasmzs a u a m w M s s Thzrzfnr z ME I Whn r s wrung wwh u praaf 7 Cawmn II Evzry mam mg z has m 5mm mznsurz as b mu wsznng z A m me rmhn g w AREDnMIMnnswnpnm z m nmhwrmnrg n m hath w mm m Hump w mmmwmwp Wm M Mm 3v y quotum wmmaw w lt17 H mm mpwwww a mum wuqaav mwwmg M H WW9 a w quotum quotWm m sq m w 1 av y mum NNMM am an x w lt17 H mm u 5W M W m4 3 WA a mu Mnmv WW Wm m I v amuzsmm 517 MW zums M saw mm mm Am H WW9 a akusmq unawmmww lt17 H mm mmwwm saw mm mm Am H umnmg HGGZ 5 Z H WW9 a M av a m Wm mm W Mudwwrmawmpww awuuuduwwwap Wm w raw 7 y mum nymwadwdamaqu H WW9 a M av 43 m Wm mm a y mm m Hump w mmmwmwp Wm M Mm unmmmww H umnmg HGGZ 5 Z c1oo dsmm vm Euon sx Jon mupan mm 2x apwmmmm u H WW9 aoozsz mm mpr mm m qvveqm a may 2 H WW9 wv uH w mpervmus Ta ransan Axmm SysTems E m me pm 1 rm 1 a 5m in mm mm m Mn mesmqu mm m w asvruw mMhngm ws uw msm h Harman mm man m o mst Wpdmdu d quotmm mm mm M mhmmm mama s uchd s Axmms wmqmmwmmm x mm mmmm w m 252003 252001 Euchd scnmmnn Nnhms 1 1m 1 twist d ulqud huMsnlqui 1 mthsmmwmqmmnmmsmm 1 11m ummwmmwmamhrmlqh um 5 mummmmmm Euchd sAxmmerndem Versmn 1 mm my a man 1 m m pnxxw mm D Md 0 H Euchd sAxmmerndem Versmn 1 Falwer MPale rpMQnmlquAQuP hrms s 111mmmpmmwmq z mmmMumwwmmnm unqpunEunhrAksumwmhsbnmm111E111 wmnmwmmz I z 39 n 1 252003 Euchd sAxmmerndem Versmn nmpmmmpummmnvmms wwmmmmwm nmmmm quotwwmnmmm m punmmsmmmma mi m wmnmwmmz rummu numb mlquhu lmsa Mm umumm m Euchd sAxmmerndem v m e 1 myquot ad wpunQn lquA uPQhrms s wwmmmmwmq 1 ml mmwwmmmm x a Falwrvpum mlwwpu Mm mum AI quotmm m M mlquhu lmsa m wmmam Euchd sAxmmerndem Versmn 1 FalerF MPmIWWpu QnmlquA uP hrIns s wwmmmmwmq mmmmmwmwmmm mpunmmsmmmmamm mwmwm as a Falerpxm mlwrypun mlquhu hmsa Mummumme ummmmmmm 5 humn mhnwwpuumuthrms sa mu mmww s mm Wm mm 252003 Euchd s Axmms r Pmb ems Werem pm 1 EuAdMIsbVumssmhnslpmmun ugmhd rlmsvmrg lpuprhswmhwsmarymmsu was a mmmmwusasmuwmmm nmmnmummmwm mummmmmwmmmm o m mmmmmmm s Myaumnmmmmmm Memmm Axmm Sysums Mm mm mummm nmmamqummmmywmmmm mm WWW m m znnumnumvmmu my lvw n mm W Anms WWW mm SMS Axmm Sysmn mm m a mu Wm wry m mmmm Mm mym nmi mm m quot quot M quot mm Mm m mm m mmmm Yuuvpumdmmmnmns x y n Wm awnunnv aaunm umsrm Qua mun nummmmmmnsmwm mmmnmmmmmm Wm swarm awpmmmmmMmmwm 5 mm m mmmgu 6Axmm Sysmn mnmpuns quotmg m m m wwxmm m Mmmgm mum m Mm mm Wmmmmmm M Mum wwmwmnw 252003 6Axmm5 Hm mm W mwmmwhm mm ammummmmmm mmmlm Mm n mm m mm s mm mm mmmi a g n inomm u m quot anmmmm Magnumquot mm M hwmtmmm mums quotmamamm mm M wwwmwwwm H mm mu wwwman hmsnxm ymMrAqumNsumhmnmkw smss Axmm Sysmn mm B Myrwmumwn ms mum mnum w mum m4mhm4m m Wm M m ah my gtmywp nm mm 11 5459mm am a 1 mm mm hummum mmmwmmxm m mmnm mm mm m ammmm m mem 3 my mm quot mtmmme agwnlxtnd M m mmw mh mwmm mm Wmmm m PmMmllelf 1171 mm 5 mm mm fnm39 m 3 5M56Axmm5ysum mm mm M my quotgmaw Wmmmmmwux mmmmmnmawm gt my mama nad sbml mm m pm on mm Ntntkugwhtkummmwm mm mm m 5 uMMI ymmwnhm m mmzz tuknqumwmusms m m n mwmmmmmmsmm pumm mgmmwmmmmmMumman m m mmmmmwmum manng and mr cm 252003 Whn r s The CENTER 0f 1 mang e WM f rr s nmqmwmb 252003 Circumcemer Cemmid 24 Onh ocemer 252003 Incenter Veg 1 The 4Cemers so for L Ceva39s Theorem zszooa w my mm m n m mm m mom a m mm m mspm m Ceva39s Theorem n d c n me am Werszmng nppnsnz smzs m m n m w M mannw A ANBLCM 1 Ceva39s Theorem Ceva39s Theorem BL may Hquot mm 4mm KAPBL KAABP zszooa Ceva39s Theorem m KABM 4mm 7mm arm A W mm 7K Ceva39s Theorem K WWW 1mm MW ANV AM NBquot KmBCNmeBPN Ceva39s Theorem ANBLCMVKAMQKAABP mm 7 NW 39 mm mm mm 39 zszooa Ceva39s Theorem an assure mm ANBLt M m Ward AL T mmmnwum mm to mmmngAkm N N dmmm m N p o 5mm N mosnmmmz N u MA mm mm hwn hzsws w mm mm w AN Ceva39s Theorem N W Medians ampCemroid 252003 Pm 5m L M m N m mmpmmm mm n v mamrmmmr Nauw mmm 1 wmu wmc Mn 1 AJL wuwu MA AR Angle Bisectors amp Incenter ck m AN mm WNRLEVAE wa Inr39adius amp Area 252003 mm y Heron39s Formula BVBT gtlt Y cm 5ch m Heron39s Formula 1 pm 252003 Heron39s Formula Heron39s Formula mm ymu mm mm Lyhmrnw an AV Heron39s Formula swan mm mm nxm mm wsm mngwa mgwqglgwma y 9 ZSZDDE Heron39s Formula 7 Ema41 Nan m Am AIZA mm 5n EVAV m2 em rz yL 7n Heron39s Formula x alame 9n mm LZBI oLIKxokaEoLEav m LZBI ukx mdLle E L avg zszooa Heron39s Formula 9 l 7w M 7 mm m nunmmwm mu man Heron39s Formula In KlMRE JXanXXth mm m 39xranthLVr K lxranthLJIlml K lyanthLJlK KlnlxranthL Chapter 10 Hypercycles and Horocycles There is a curve peculiar to hyperbolic geometry called the horocycle Consider two limiting parallel lines 6 and m with a common direction say 9 Let P be a point on one of these lines P E Z If there exists a point Q E m such that the singly asymptotic triangle APQQ has the property that APQQ E AQPQ then we say that Q corresponds to P If the singly asymptotic triangle APQQ has the above property we shall say that it is equiangular Note that it is obvious from the de nition that if Q corresponds to P then P corresponds to Q The points P and Q are called a pair of corresponding points Theorem 101 If points P and Q lie on two limiting parallel lines in the direction of the ideal point 9 they are corresponding points on these lines if and only if the perpendicular bisector of PQ is limiting parallel to the lines in the direction of 9 Theorem 102 Giuen any two limiting parallel lines there exists a line each of whose points is equidistant from them The line is limiting parallel to them in their common direction PROOF Let t and m be limiting parallel lines with common direction 9 Let A E l and B E m The bisector of ABAQ in the singly asymptotic triangle AABQ meets side B9 in a point X and the bisector of LABS meets side AX of the triangle AABX in a point C Thus the bisectors of the angles of the singly asymptotic triangle AABQ meet in a point C Drop perpendiculars from C to each of l and m say P and Q respectively By Hypothesis Angle ACAP E ACAM M is the midpoint of AB and ACBQ E ACBM Thus CP E CM E CQ Thus by SAS for singly asymptotic triangles we have that ACPQ E ACQQ gt and thus the angles at C are congruent Now consider the line CS2 and let F be any point on it other than C By SAS we have ACPF E ACQF If S and T are the feet of F in l and m then we get that APSF E AQTF and FS E FT Thus every point on the line CS2 is equidistant from t and m I This line is called the equidistant line 87 Theorem 103 Giuen any point on one of two limiting parallel lines there is a unique point on the other which corresponds to it Theorem 104 If three points P Q and R lie on three parallels in the same direction so that P and Q are corresponding points on their parallels and Q and R are corresponding points on theirs then P Q and R are noncollinear Theorem 105 If three points P Q and R lie on three parallels in the same direction so that P and Q are corresponding points on their parallels and Q and R are corresponding points on theirs then P and R are corresponding points on their parallels Consider any line 6 any point P E l and an ideal point in one direction of 6 say 9 On each line parallel to l in the direction 9 there is a unique point Q that corresponds to P The set consisting of P and all such points Q is called a horocycle or more precisely the horocycle determined by l P and Q The lines parallel to l in the direction 9 together with l are called the radii of the horocycle Since 6 may be denoted by P9 we may regard the horocycle as determined simply by P and Q and hence call it the horocycle through P with direction 9 or in symbols the horocycle P All the points of this horocycle are mutually corresponding points by Theorem 105 so the horocycle is equally well determined by any one of them and Q In other words if Q is any point of horocycle P 9 other than P then horocycle Q Q is the same as horocycle P If however P is any point of Z other than P then horocycle P Q is different from horocycle P 9 even though they have the same direction and the same radii Such horocycles having the same direction and the same radii are called codirectional horocycles There are analogies between horocycles and circles We will mention a few Lemma 101 There is a unique horocycle with a giuen direction which passes through a giuen point There is a unique circle with a giuen center which passes through a giuen point Lemma 102 Two codirectional horocycles have no common point Two concentric circles have no common point Lemma 103 A unique radius is associated with each point of a horocycle A unique radius is associated with each point of a circle A tangent to a horocycle at a point on the horocycle is de ned to be the line through the point which is perpendicular to the radius associated with the point No line can meet a horocycle in more than two points This is a consequence of the fact that no three points of a horocycle are collinear inasmuch as it is a set of mutually corresponding points cf Theorem 104 Theorem 106 The tangent at any point A of a horocycle meets the horocycle only in A Euery other line through A except the radius meets the horocycle in one further point B fa is the acute angle between this line and the radius then dAB is twice the segment which corresponds to 04 as angle of parallelism MATH 6118 090 Spring 2004 88 CHAPTER 10 HYPERCYCLES AND HOROCYCLES PROOF Let t be the tangent to the horocycle at A and let 9 be the direction of the horocycle If 25 met the horocycle in another point B we would have a singly asymptotic triangle with two right angles since A and B are corresponding points In fact the entire horocycle except for A lies on the same side oft namely the side containing the ray AQ Let k be any line through A other than the tangent or radius We need to show that k meets the horocycle in some other point Let 04 be the acute angle between k and the ray AQ Let C be the point of k on the side oft containing the horocycle such that AC is a segment corresponding to 04 as angle of parallelism RECALL e d tana2 The line perpendicular to k at C is then parallel to A9 in the direction 9 Let B be the point of k such that C is the midpoint of AB The singly asymptotic triangles AACQ and ABCQ are congruent Hence ACBQ a B corresponds to A and B E A I A chord of a horocycle is a segment joining two points of the horocycle Theorem 107 The line which bisects a chord of a horocycle at right angles is a radius of the horocycle We can Visualize a horocycle in the Poincare model as follows Let 6 be the diameter of the unit disk whose interior represents the hyperbolic plane and let 0 be the origin It is a fact that the hyperbolic circle with hyperbolic center P is represented by a Euclidean circle whose Euclidean center R lies between P and A As P recedes from A towards the ideal point 9 R is pulled up to the Euclidean midpoint of QA so that the horocycle A Q is a Euclidean circle tangent to the unit disk at Q and tangent to l at A It can be shown that all horocycles are represented in the Poincare model by Euclidean circles inside the unit disk and tangent to boundary circle For the Poincare upper half plane model our horocycles will be circles that are tangent to the z axis Figure 101 H1 and H2 are horocycles in the Poincare model 01 is a circle Another curve found speci cally in the hyperbolic plane and nowhere else is the equidis tant curve or hypercycle Given a line 6 and a point P not on 6 consider the set of all points Q on one side of 6 so that the perpendicular distance from Q to l is the same as the perpendicular distance from P to l MATH 6118 090 Spring 2004 89 The line 6 is called the axis or base line and the common length of the perpendicular segments is called the distance The perpendicular segments de ning the hypercycle are called its radii The following statements about hypercycles are to t t about regular Euclidean circles H Hypercycles with equal distances are congruent those with unequal distances are not Circles with equal radii are congruent those with unequal radii are not to A line cannot cut a hypercycle in more than two points 03 If a line cuts a hypercycle in one point it will cut it in a second unless it is tangent to the curve or parallel to it base line F A tangent line to a hypercycle is de ned to be the line perpendicular to the radius at that point Since the tangent line and the base line have a common perpendicular they must be hyperparallel This perpendicular segment is the shortest distance between the two lines Thus each point on the tangent line must be at a greater perpendicular distance from the base line than the corresponding point on the hypercycle Thus the hypercycle can intersect the hypercycle in only one point 01 A line perpendicular to a chord of a hypercycle at its midpoint is a radius and it bisects the arc subtended by the chord a Two hypercycles intersect in at most two points 1 No three points of a hypercycle are collinear In the Poincare disk model let 9 and A be the ideal end points of t It can be shown that the hypercycle to 6 through 9 is represented by the arc of the Euclidean circle passing through A B and 9 This curve is orthogonal to all Poincare lines perpendicular to the line 6 In the Poincare upper half plane model the hypercycle will be represented by an arc of a Euclidean circle passing through A B and Q In the Poincare disk model 9 a Euclidean circle represents a a hyperbolic circle if it is entirely inside the unit disk A b a horocycle if it is inside the unit disk except for one point where it is tangent to the unit disk c an equidistant curve if it cuts the unit disk non orthogonally in two points d a hyperbolic line if it cuts the unit disk orthogonally A similar situation is true for the Poincare upper half plane model if It follows that in the hyperbolic plane three non collinear points lie either on a circle a horocycle or a hypercycle accordingly as the perpendicular bisectors of the triangle are concurrent in an ordinary point an ideal point or an ultra ideal point MATH 6118 090 Spring 2004 MATH 6118 I was going To Try powdered wa rer39 bu r I quot 39n39T Know whaT To add Undefined Terms Line DisTance Ange Measure Axioms Exis rence Axiom The collec rion of all oin rs forms a nonem r se r There is more Than one poin r in Tho r se r Incidence Axiom Ever39y line is a se r of poin rs For39 every pair of dis rinc r poin rs and B There is exac rly one line 8 con raining A and B Defini rion Two lines 8 and m are garallel if They do no r i n rer39sec r 39 eor39em If 8 and m are Two dis rinc r nonparallel lines Then There is exac rly DrlnnF DisTonce 3 Ruler Axiom For every pair of oin rs P and There is a real number dPQ For each line 8 There is a 1 1 rnrrpcnnnripnrp 39Frnm P n D n nn P and Q correspond ro xy real numbers LI llh Q IX 11 Feb2008 MATH 6118 Defini rions C is be rween A and B ABC if C lies on The line AB and efini rion C is be rween A and B ABC if C lies on The line AB and ABABUP ABC lA BABUPABI39 DefiniTions The leng rh of segmen r AB is d A B Two se men rs are con r39uen r if They have The same leng rh I L m u TL n l dPO3 2 0mm dPQ 2 O Ihen Plane SeparaTion For39 every line 8 The poinTs ThaT do noT lie on 8 form Two dis oinT nonem T seTs H1 and H2 so ThaT H1 and H2 ar39e convex ii If P is in H1 and Q is in H2 Then PQ 11 Feb2008 MATH 6118 DefiniTion a line 8 and exTer39nal poinTs AB A and R are on The same side if AB does noT i nTer39secT 8 A and B are on opgosiTe 4 of f if AB does inTer secT 8 An angle is The union of Two nonopposiTe r39ays sharing The same endpoinT Tlna innFanian nF Ina nnnla ic Ina i nTer39secTion of Two half planes DefiniTion A B C are noncollinear poinTs The Trianqle ABC is The union of The Three segmenTs AB BC and AC 11 Feb2008 MATH 6118 POSLH39S Theorem AABC be a Triangle and f a line so Tha r none of A B and C are on 9 If 8 i n rer39sec rs AB Then 8 i n rer39sec rs eiTher AF nr RF Proof Angle Measure Pr39o rr39ac ror39 Axiom For39 every an e BAC There is a number m BAC so ThaT i O s mBAC s 180 ii mBAC 0 iff AB AC measure r39 on ei rher39 side of a line if AD is be rween AB and AC mBAh mhAC mBAC 11 Feb2008 MATH 6118 Be rweenness func rion o 1 1 correspondence fz 9Rso Tho rdP fPfQ Theorem If ABC lie on 8 Then C is w A n iff fAltfcltfBor fA gt fC gt fB If ABC on 8 The exoc rly one of Them lies between The other l39wo Be rweenness heor39em LeT 8 be a line and AD on f If B and E on opposi re sides of 8 Then rays AB and DE do no r i m er39sec r Be rweenness heor39em Each angle has a unique bisec ror39 Given AABC Ie r D be 4A 4 r39 The ray AD mus r i n rer39sec r BC Triangle Congruency Two Triangles are congruen r if There is a 1 1 correspondence be rween The ver rices so Tha r The corresponding irlpc map rnnnrnpn l39 nnrl rnrrpcnnndinn angles are congruen r xiom Given AABC and ADEF so Tha r AB 3 DE BC 3 EF and ABC 3 DEF Neu rr39al Geome rry Resul rs heor39em ASA Given AABC and ADEF so Tha r CAB z FDE AB 2 DE and ABC 8 DEF Then AABC z ADEF In AABC if ABC 3 ACB Then 11 Feb2008 MATH 6118 18 Neu rral Geome rry Resul rs heorem ExisTence of Perpendiculars Given line 8 and P no r on 8 There exis rs a line m Through P Tha r is perpendicular n P AlTernaTe InTer39ior Angles heor39em AlTernaTe InTer39ior Angles Theorem If Two lines are cuT b o Transversal so ThaT There is a pair of rnnnrnpn l39 nl l39prnn l39p in l39prinr rmnlpcl Then The Two lines are parallel Existence of Parallel Lines heor39em If m and n are dis rinc r lines bo rh er39 endicular39 To 8 Then m and n are parallel 11 Feb2008 MATH 6118 21 Uniqueness of Perpendiculars heor39em If P is no r on 8 Then The er39 endicular39 dr39o ed from P To 8 is unique 11 Feb2008 MATH 6118 22 Ex rerior Angle Theorem heor39em An ex rer39ior angle of a Triangle is qr39ea rer39 Than ei rher39 r39emo re i n rer39ior39 angle 11 Feb2008 MATH 6118 23 Angle Angle Side Cr39i rerion heor39em Given AABC and ADEF so Tha r AC 3 DP BAG 3 EDF and ABC 3 DEF Then AABC z ADEF 11 Feb2008 MATH 6118 24 HypoTenuse Leg Cri rerion heor39em Two righ r Triangles are con want if The h o renuse and la of one are congruen r respec rively To The Invnn l39pmlcp and Ian n39F Hap VIhon Side Side Side Cr39i rerion heor39em Given AABC and ADEF so Tha r AC 3 DF AB 3 DE and BC 3 EF Then AABC quot39 ADEF 11 Feb2008 MATH 6118 26 Saccheri Legendre Theorem The sum of The measures of any rwo an les of a rr39ian e is less Than 180 Lemma If A B and C are noncollinear Jl IA I Innl Inr39l Saccheri Legendre Theorem heor39em Saccher39iLegendre The sum of The an les I an rr39ian Ie is less Than or39 equal To 180 1 If A B and C are noncollinear Then IACI lt IABI IBCI DefecT of a Triangle The defecT of a Triangle AABC is The number heorem AddiTiviTv of Defech LeT AABC be any Triangle and D lie an AB lnan39 defAA BC defAACD defABCD Qualify of Defec r heor39em n If There exis rs a Triangle of defec r 0 Then a r39ec rangle exis rs 39 I a r39ec rangle exisTs Then every Triangle has defec r O QualiTy of Defec r h of Proof AARC hnlt defer O There is a righ l39 Triangle wiTh defec r O 9 we can MMJ quot quot 9 we can cons rruc r arbi rrarily large rec rangles 9 ll righ r Triangles have defec r O 9 all Triangles have defec r O PosiTive DefecT Marx If There is a Triangle wiTh osiTive defecT Then all Tr39ian les have posiTive defecT 11 Feb2008 MATH 6118 32 MATH 6118090 NonEuclidean Geometry Exercise Set 3 Solutions 1 Prove that an equiangular triangle is equilateral allsides are congruent By the Converse t0 the Isosceles triangle theorem if in AABC we have that 4A 5 LB then BC E AC Since the triangle is equiangular we also have that 4A E AC from which it follows that BC E AB Thus we have that AB E BC E AC and the triangle is equilateral k In triangle AABC below let A and C39 be the midpoints ofBC andAB respectively Let AA and CC39 intersect at G Suppose thatAG has length 2 and suppose that the circle through A A and C bisects the segment C39G What is the length ofCG B AUG We know that since A and C39 are midpoints the line AC is parallel to Thus by the Converse t0 the Alternate Interior Angles Theorem LA39C39GE LACG and AC39A39G E CAG By vertical angles AC39GA39 E ACGA s0 AA39C39G AACG Thus Qz j cngg z 2 2C39G24EG AG A39G AG 1 where E is the point of intersection of CG and the circle which by hypothesis is the midpoint From our knowledge of circles we have that since G is the point of intersection of the chords AA and CE then Exercise Set 1 Page 1 Spring 2008 EGCG A39GAG iCGCG 1X 2 4 CG28 CGJ 2JE 3 In triangle AABC given that the medians AA and BB intersect at right angles and that a 3 and b 4 whatis c Let X denote the length of the median AA and let y denote the length of the median BB The centroid G divides the medians as demonstrated in the figure to the right By applying the Pythagorean Theorem to AAGB39 AA39BG and AABG we get era allellef Solving we get that x2 334 or X J Thus yJ and c22xj 5 3 3 3 2 3 So cJE In a triangle AABC suppose that AABC 60 a 5 and C 8 What is b By the Law of Cosines b2 a2 c2 2ac cos60 25 64 25812 49 4 So b 7 Whatis thearea ofAABC ifa3b5andc6 356 9 By Heron s Formula s 7 and 2 K 7x7 3x7 5x7 6562J Whatis the area ofthe incircle of AABC ifa 5 b 6 and c 7 9 The area of AABC is given by K rs Where r is the radius of the incircle and s is the semiperimeter We nd the area by Heron s Formula and then 0 Exercise Set 1 Page 2 Spring 2008 K 455 as bs c 72 2 8 8 99 59 69 7 9 2 So the area of the circle is 1427172281 3 Exercise Set 1 Page 3 Spring 2008 gtgtgtL hi n58 9h 15 938 veryone IS a foal for IT leaST Tlve minu res a day WISDOM consis rs of no r ding The limi r 28Jan 2008 MATH 6118 There exis r rwo ir39r39a rional numbers a and b so Tha r ab is r a rional Than power39 cor39r39ust buT we s rill need elec rr39ici ry Ts Absolu rel Theorem Every Triangle is isosceles C A 1 LC Call if I l is no r perpendicular To AB Trian e is Isosceles Theorem Every Triangle is C isosceles A 1 LC Call if I is no r perpendicular ro AB 9 Cons rruc r The perpendicular bisec ror of AB Call if m and Ie r A quot AB n m D Trian e is Isosceles Theorem Every Triangle is C isosceles 1 LC Call if I is no r perpendicular ro AB 9 Cons rruc r The perpendicular bisec ror of AB Call if m and Ie r quot AB n m 3 Inm Cai rP A 1 Tr39ian la is Isosceles Cons rr39uc r The bisec ror39 c of LC Call if I is no r of AB Call if m and Ie r D AB n m gt mm P To BC and Tram P To AC Tr39ian la is Isosceles 1 Cons rr39uc r The bisec ror39 c of LC Call if I is no r of AB Call if m and Ie r D AB n m A E B mm P To BC and Tr39om I F P To AC 5 LACP E L BCP P 28Jan 2008 MATH 6118 9 Tr39ian la is Isosceles Cons rr39uc r The c er39 endicular39 bisec ror39 of AB Call if m and IeT D AB n m Inm Cai rP Drop a perpendicular mm P To BC and mm P To AC A LACP s 4 BCP E B EP 2 4 CFP as F They are bo rh r39igh r anales 28Jan 2008 MATH 6118 10 Tr39ian la is Isosceles Cons rr39uc r The c er39 endicular39 bisec ror39 of AB Call if m and IeT D AB n m Inm Cai rP Drop a perpendicular mm P To BC and mm P To AC A LACP s 4 BCP E B EP 2 A CFP F ACEP s ACFP by AAS 9 4 Tr39ian la is Isosceles Inm Cai139P 6 Drop a perpendicular from P To BC and from P To AC LACP E A BCP LCEP s 4 CFP ACEP s ACFP A EPE FPand CEECF E by CPCTC 9 09 4Fquot Tr39ian la is Isosceles Inm Cai139P Drop a perpendicular from P To BC and from P To AC LACP E A BCP LCEP s 4 CFP ACEP s ACFP EP 2 PP and CE 2 CF LAEP E L BFP as They are bo rh r39igh r angles C A 9 10 CLAIM EA 2 FB Tr39ian la is Isosceles Inm Cai139P 6 Drop a perpendicular from P To BC and from P To AC LACP E A BCP LCEP s 4 CFP 6 quot ACEP E ACFP A 8 9 EP E PP and CE E CF E LAEP E L BFP
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'