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Fund Concepts of Geometry

by: Felicita Lockman

Fund Concepts of Geometry MATH 3181

Felicita Lockman
GPA 3.68


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This 8 page Class Notes was uploaded by Felicita Lockman on Sunday October 25, 2015. The Class Notes belongs to MATH 3181 at University of North Carolina - Charlotte taught by Staff in Fall. Since its upload, it has received 59 views. For similar materials see /class/228920/math-3181-university-of-north-carolina-charlotte in Mathematics (M) at University of North Carolina - Charlotte.

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Date Created: 10/25/15
Hilbert s Axioms of Geometry and Consequences Incidence Geometry We are now ready to begin our study of geometries in earnest We will study neutral geometry based on the axioms of Hilbert This means that we will study all that we can almost without the introduction of a Parallel Postulate of any sort At the appropriate time we will add a parallel postulate and see where we will be led For an ease of notation let ABC denote the statement that the point B lies between the point A and the point C DEFINITION Let t be any line and let A and B be any points that do not lie on t If A B or if the segment AB contains no point lying on t we say thatA and B are on the same side of t If A i B and ifAB contains a point of we say thatA andB are on opposite sides of l Let us quickly reView the incidence axioms AXIOM 1 1 For every point P and for every point Q not equal to P there exists a unique line that passes through P and Q AXIOM 1 2 For every line there exist at least two distinct points incident with t AXIOM 1 3 There exist three distinct points with the property that no line is incident with all three of them This does not seem like much but already we can prove several easy properties that any set satisfying these three axioms must have PROPOSITION 1 If t and m are distinct lines that are not parallel then f and m have a unique point in common Proof Since and m are not parallel and since 7 m they must have at least one point in common Assume that they have more than one point in common They then have at least two points in common Axiom I 1 says that two points determine a unique line so m which is contrary to the hypothesis Thus and m have a unique point in common DEFINITION Two or more lines are concurrent if they intersect in one common point DEFINITION Two or more points are collinear if they are all incident with the same line We have four other results to mention PROPOSITION 2 For every line there is at least one point not incident with it PROPOSITION 3 For every point there is at least one line not incident with it PROPOSITION 4 For every point there exists at least two lines incident with it PROPOSITION 5 There exist three distinct lines that are not concurrent Betweenness Axioms AXIOM B l If AB C then A B anal C are three distinct points all lying on the same line anal C B 1 A AXIOM B 2 Given any two alistinct points B anal D there exists points A C anal E lying on as such that ABD BCD and BDE Axiom B 3 If A B anal C are three distinct point lying on the same line then one anal only one of the points is between the other two PROPOSITION 6 For any two pointsA anal B gt gt z ABmBAAB n uazlA B C A 5 Figure 1 It seems clear from Figure 1 that every point P lying on the line ll through A B and C must gt gt either belong to ray AB or to an opposite ray AC This statement seems similar to the second assertion of Proposition 6 but it is actually much more complicated You are now discussing four points and not the three of Proposition 6 Let us call the assertion gt CAB anal apointP is collinear with A B anal C implies that P e ABU AC the line separation property This is something that you will prove but knowing that it can be proven we shall use it as we need Axiom B 4 Plane Separation Axiom For every line anal for any three points A B anal C not lying on is i IfA analB are on the same siale of analB anal C are on the same siale of z thenA anal C are on the same siale of l ii IfA anal B are on opposite siales of anal B anal C are on opposite siales of thenA anal C are on the same siale of l Recall the de nitions of same siale and opposite siales Note that this axiom indirectly guarantees that our geometry will be only twodimensional for such an axiom will not hold in three dimensional space The pointsA and C may lie in the same plane with l yet the pointB need not In this case the conditions would be satis ed yet the conclusion would not hold DEFINITION Let ll be any line A side of the line ll containingA is the set of all points that are on the same siale of l as A This is also calleal the half plane bounded by l containing A PROPOSITION 7 Every line bounds exactly two half planes and these half planes have no point in common PROPOSITION 8 Given ABC and ACD then BCD and ABD PROPOSITION 9 Line Separation Property If C AB and l is the line through A B and C then for every pointP incident with z P is incident with either ray A B or the opposite ray A C The next theorem is very interesting Pasch39s Theorem If AABC is any triangle and l is any line intersecting side AB in a point betweenA and B then also intersects either side AC or side BC IfC is not incident with then does not intersect both AC and BC Figure 2 Proof Compare Figure 2 Either C 6 or C If C 6 then we are done Thus let us assume that C E l By hypothesis AB e l and mAB 2 Q Thus A and B are on opposite sides of Then since there are only two sides to any line either C is on the same side of l as A or it is on the opposite side HA and C are on the same side of l thenB and C are on opposite sides of Thus mBC i Q Likewise ifA and C are on opposite sides then mAC i Q and we are done I The following propositions are le to you to prove PROPOSITION 10 Given ABC then AC ABUBC andB is the only point common to the segments AB and BC PROPOSITION 11 Given ABC then B is the only point common to the rays E and E and A B A DEFINITION Given an angle ACAB we shall say that a pointD is in the interior of the angle if D is on the same side oflA B as C and on the same side ofA C as B Note that this implies that the interior of an angle is the intersection of two half planes PROPOSITION 12 Let ACAB be any angle and letD be anypoint incident with line D is in the interior of ACAB ifand only ifBDC PROPOSITION 13 If D is in the interior of angle ACAB then39 i so is every other point on the ray E exceptA ii no point on the opposite ray to E is in the interior of ACAB iiiif CAE thenB is in the interior of ADAE gt gt gt gt DEFINITION Ray AD is between rays AC anal AB if AB anal AC are not opposite rays and D is in the interior of ACAB gt gt gt gt CROSSBAR THEOREM If AD is between AB anal AC then AD intersects the segment BC Proof We are glven a ray AD between rays AB and AC Let us use proof by contrad1ctlon and assume that A DmBC Q Let AF be the oppos1te ray to AD If AFmBC P then BPC and by Propos1tlon 12 we have that P lies in the interior of ACAB However this contradicts ii of Proposition 13 which says that no point on the opposite ray can be interior to the angle Thus we have that A FmBC Q Now this means that A Dm BC Q by Proposition 6 part ii It follows that B gt and C are on the same side of the line AD Let E be a point on the line lA C so that CAE Then C and E are on opposite sides of A D and by Betweenness Axiom 4 B and E are on opposite sides of Applying Proposition 13 part iii to this situation we know that B is in the interior of ADAE which means that B and E gt are on the same side of the line AD We now have a contradiction gt Thus we have that AD 0 BC 72 Q I Lastly we need to consider the incidence of rays and triangles DEFINITION The interior of a triangle is the intersection of the interiors of the three angles A point is in the exterior of the triangle if it is not in the interior or on any of the three sides PROPOSITION 14 gt i If a ray PQ emanating from a point exterior to AABC intersects side AB in a point betweenA and B then also intersects AC or BC ii If a ray emanates from an interior point of AABC then it intersects on of the sides and if it does not pass through a vertex it intersects only one side Congruence Theorems Axiom C l If A and B are distinct points and if A39 is any point then for each ray r emanating from A39 there is a unique pointB39 on r such that B i A and AB 5 A B Axiom C Z If AB 5 CD and AB 5 EF then CD 5 EF Moreover every segment is congruent to itself Axiom C 3 IfABC A B C AB EA B and BCEB C then ACEA C Axiom C 4 Given any ABAC and given any ray A B emanating from a pointA39 then there is a unique ray AC on a given side ofline A B such that ABAC E AB A C Axiom C S If 4A 5 AB and AB 5 AC then 4A 5 4C Moreover every angle is congruent to itself Axiom C 6 SAS If two sides and the included angle of one triangle are congruent respectively to two sides and the included angle of another triangle then the two triangles are congruent According to our development congruence has been a relationship between segments or a relationship between angles In Geometry we are accustomed to seeing congruence as a relationship between triangles We can make this so by a de nition DEFINITION Two triangles are congruent if there exists a one to one correspondence between their vertices so that the corresponding sides and corresponding angles are congruent PROPOSITION 15 If in AABC we have AB 2 AC then AB 2 4C Proof This simple proof is due to Pappus Consider the correspondence of vertices A gtA B gt C and C gt B Under this correspondence two sides and the included angle of AABC are congruent respectively to the corresponding sides and included angle of AACB Hence by SAS the triangles are congruent Therefore the corresponding angles are congruent AB 5 AC PROPOSITION 16 Segment Subtraction IfABC DEF AB 5 DE and AC EDF then BC E EF Proof From what we are given let us assume that BC 1 EF By Axiom C l there exists a unique point G on ray E so that BC EEG By our hypothesis we have that G F Since AB EDE and BC EEG by Axiom C 3 AC EDG But then by Axiom C Z we have that DF 5 DG It follows from Axiom C l that G F We have reached a contradiction Thus it must be that BC 5 EF and we are done I PROPOSITION 17 Given AC 5 DE then for any pointB between A and C there exists a unique point E DEF such that AB 5 DE We can use this result to help us establish a partial ordering on the line segments in the plane DEFINITION AB lt CD or CD gt AB means that there exists a pointE between C anal D such that AB E CE PROPOSITION 18 Segment Ordering i exactly one of the following conditions holds trichotomy AB lt CD ABECD orAB gt CD ii i AB lt CD anal CD EEF thenAB ltEF iiiifAB gt CD anal CD EEF thenAB gt EF iv ifAB lt CD anal CD lt EF thenAB lt EF PROPOSITION 19 Supplements of congruent angles are congruent PROPOSITION 20 1 Vertical angles are congruent to each other ii An angle congruent to a right angle is a right angle PROPOSITION 21 For every line and every point P there exists a line through P perpendicular to Proof Either P e l or P 65 l First let us assume that P 65 l Let AB e Such points exist by Axiom 1 2 The ray E lies on one side of l By Axiom C 4 there exists a ray A X on the opposite side of fromP so that AXAB E APAB By Axiom C 1 there is a point P e A X so that AP 5 AP Since P and P39 are on opposite sides of z the segment PP39 intersects the line Let PP m If Q A then APQB 413143 EP AB AP QB Thus Wit If Q i A then APAQ E AP AQ by SAS Thus from the de nition of congruent triangles APQA AP QA and Wii Now if P e l there is an X 65 l From this point X apply the preVious technique to construct a perpendicular line to throughX By Axiom C 4 we can copy this angle on one side of l at P From the second part of Proposition 20 the other side of this angle is part of a line through P perpendicular to l I PROPOSITION 22 ASA Given AABC anal ADEF with 4A 2 AD 4C 2 AF anal AC EDF Then AABC E ADEF PROPOSITION 23 If in AABC we have that AB 5 AC then AB 5 AC anal AABC is isosceles gt gt gt gt gt PROPOSITION 24 Angle Addition Given BG between BA anal BC EH between ED anal E ACBG E AFEH anal AGBA E AHED Then zABC E ADEF gt gt gt gt gt PROPOSITION 25 Angle Subtraction Given BG between BA anal BC EH between ED anal E ACBG E AFEH anal zABC E ADEF then AABG E ADEH gt gt gt Proof Since BG hes between BA and BC we may apply the Crossbar Theorem to nd that E intersects AC Without loss of generality we may assume that this point of intersection is in fact the point G Then we have that AGC Assume that the points D F and H are chosen so that AB 5 ED BG 5 EH and BC 5 EF This is nothing but a relabeling of the points gt Let us assume then that ZABG ADEH Then there exists a unique ray EP on the same side gt gt of EH as ED so that AHEP E AABG By our assumption EP ED so by Propos1tion 24 ZABC E APEF Since ZABC E ADEF the uniqueness of E3 implies that E3 E a contradiction Thus ZABG E ADEH I As with line segments there is a natural method for defining an ordering on angles gt gt DEFINITION ZABC lt ADEF means that there ex1sts a ray EG between ED and EF so that AABC E AGEF This gives us the following results which are completely analogous to those for segments PROPOSITION 25 Ordering of Angles i Exactly one of the following conditions holds APltAQ APEAQ or APgtAQ ii If4PltZQ anal ZQEAR then APltAR iiiIfAPgtZQ anal ZQEAR then APgtAR iv If4PltZQ anal ZQltAR then 4PltAR PROPOSITION 26 SSS Given triangles AABC anal ADEF If AB 5 DE BC 5 EF anal AC EEF then AABCE ADEF PROPOSITION 27 Euclid39s Fourth Postulate All right angles are congruent to each other Axioms 0f Continuity These axioms are the axioms which give us our correspondence between the real line and a Euclidean line These are necessary to guarantee that our Euclidean plane is em complete The first axiom gives us some information about the relative sizes of segments as compared one to another ARCHIMEDES39 AXIOM If AB anal CD are any segments then there is a number n such that if segment CD is laid of n times on the ray ZB emanating from A then a point E is reached where n CD 5 AE analB is betweenA analE This is derived from the Archimedean Axiom in the real number system This should not be surprising for we wish to have a onetoone correspondence between each euclidean line and the set of real numbers In the real line the Archimedean Postulate takes on the avor Archimedean Postulate Let M anal e be any two positive numbers Then there is a positive integer n such that ne gt M The main point for geometry is that if you choose any segment to be of unit length then every segment has nite length with respect to this measure Nothing can be too big nr CD gt AB and nothing can be too small n r EF gt CD where CD was chosen as our unit length This is still not enough for the purposes of geometry for the set of rational numbers Q satis es this property but causes trouble in another situation Let us consider the set of rational points in the cartesian plane Call this Q X Q Consider the ray passing from the origin through the point 11 This segment has length 15 Now on the ray making up the positive real axis we are unable to nd a point satisfying Congruence Axiom 1 No point exists whose distance to the origin is wE We require a stronger property Dedekind39s Axiom Suppose that the set of all points on a line is the union 21 U22 of two nonempty subsets such that no point of 21 is between two points of 22 anal vice versa Then there is a unique point O lying on f such that PI OPZ ifanal only ifP1 e 21 anal P2 6 22 anal 0 2 131132 Without Dedekind s Axiom there is no guarantee that there is a segment of length 727 or of length e or of certain other non constructible lengths eg It is Dedekind39s Axiom that allows us to make the correspondence of the line in our geometry and the real line There are welldefined geometries that exist without the Dedekind Axiom such as the geometry of the sural plane They do not have all of the properties that we wish to have or to which we are accustomed to having It is with Dedekind39s Axiom that we are able to introduce a coordinate system and do geometry analytically in the fashion of Fermat and Descartes To see why we will want Dedekind39s Axiom consider the manner in which you construct the perpendicular to a given line at a given point First using the given point as a center draw a circle of positive radius The circle intersects the line in two points At each of these points you then construct a circle of larger radius and these two circles intersect in two points Drawing the line between the two points of intersection gives a line perpendicular to the given line at the given point There are two problems with this proof and they are both very subtle Why does the line intersect the circle at all Why do the two circles then intersect In the sural plane these are not necessarily true The rst of these problems is addressed by the following principle which follows from Dedekind39s Axiom We de ne a point A to be inside a circle centered at O with radius OR if OA lt OR A point B is outside the circle if OB gt OR Elementary Continuity Principle If one enalpoint of a segment is insiale a circle anal the other outsiale then the segment intersects the circle The second problem above is that of the intersection of two circles It is addressed by the following principle which is again a corollary of Dedekind39s Axiom Circular Continuity Principle If a circle 7 has one point insiale anal one point outside another circle 7 then the two circles intersect in two points


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