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# Class Note for MATH 409 with Professor Martin at KU

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Date Created: 02/06/15

Axioms de nitions and theorems for plane geometry Math 409 Spring 2009 March 9 2009 The building blocks for a coherent mathematical system come in several kinds 0 Unde ned terms These are typically extremely simple and basic objects like point and line so simple that they resist being described in terms of simpler objectsi Every system has to have some unde ned termsiyoulve got to start somewhere But in general the fewer the better 0 PostulatesAxiomsi These are basic facts about unde ned terms The simpler and more funda mental they are the better For example every pair of points determines a line or if z y then 77 y z i 0 De nitions We can de ne new terms using things that we already knowi o Theoremsi These are the statements that makes mathematics what it isithey are facts that we prove using axioms de nitions and theorems that welve proved earlieri Propositions Lemmas Corollaries are all species of theorems 1 Unde ned terms We will start with the following unde ned terms1 0 Point 0 Line in nite straight line Angle Distance between two points 0 Measure of an angle Welll write AB for the distance between two points A and B and well write mAABC for the measure of angle ABCVi Welll be careful to distinguish between an angle which is a thing and its measure which is a number So for instance the two statements AABC AXYZ77 and mAABC mAXYZ77 donlt mean the same thingithe rst says these two angles are actually the same angle while the second just says that they have the same measurei 1Euclid included de nitions of these terms in the Elements but to a modern reader his de nitions are really intuitive explanations rather than precise mathematical de nitions For example he de ned a point as that which has no part a line as length without breadth and an angle as the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line The intuition here is that an angle records the difference between the directions of two lines meeting in a point 2 Some de nitions When writing de nitions7 it is good practice to emphasize the word you are de ningithat makes it easier on the reader and for that matter the writer as well De nition 1 A collection of three or more points is collinear if there is some line containing all those points De nition 2 Two lines are parallel if they never meet De nition 3 When two lines meet in such a way that the adjacent angles are equal7 the equal angles are called right angles7 and the lines are called perpendicular to each other De nition 4 A circle is the set of all points equally distant from a given point That point is called the center of the circle What about the term line segment We all know what that isiitls the portion of a line between two points But what does between mean With a little thought7 we can de ne between using two concepts we already have the unde ned term distance and the de nition of collinear Once welve done that7 we can de ne what a line segment is It7s important to get these two de nitions in the proper order De nition 5 Given three distinct collinear points A7 B7 C7 we say that B is between A and C if AC gt AB and AC gt BC De nition 6 The line segment AB between two points A and B consists of A and B themselves7 together with the set of all points between them 3 Axioms Axi0migt If AU B are distinct points then there is exactly one line containing both A and B which we denote AB or BA This is Euclidls rst axiom Notice that it includes a de nition of notation Also7 itls false in spheri cal geometryiif points A and B are polar opposites7 then every one of the in nitely many great circles containing A contains B as well and vice versa The next group of axioms concern distance Axiom 2 AB BA Axiom 3 AB 0 i A B The word iff is mathematicianls jargon for if and only if That is7 the axiom says that two different things are true First7 if A B7 then AB 0 Second7 if AB 07 then A B Logically7 these are two separate statements Axiom 4 If point C is between points A and B then AC BC AB Axiom 5 The triangle inequality If C is Lot between A and B then AC BC gt AB Now7 some axioms about angle measure Axiom 6 New version a mMBAC 0 i BAC are collinear and A is not between B and C b mABAC 180 BAC are collinear and A between B and C Note In an earlier version of these notes 1 said if instead of iff but in fact we want both directions of both assertions as axioms It might seem odd to start by talking about angles that arenlt really angles77 because they are de ned by three collinear pointsl On the other hand its always a good idea in mathematics to look at extreme cases These axioms make sense if you think about what happens if the points move around a little bit To understand the rst part of Axiom 6 imagine nudging B so that it is just off the segment F then mAABC should be very close to 180 and the less youlve nudged B the closer mAABC gets to 180 Axiom 7 Whenever two lines meet to make four angles the measures of those four angles add up to 360 Axiom 8 Suppose that AB C are collinear points with B between A and C and that X is not collinear with A B and C Then mAAXB mABXC mAAXC Moreover mAABX mAXBC mAABC We know that ABC 180 by Axiom 6 Axiom 8 has a bit more going on than its predecessors so herels a picture that illustrates it The rst statement says that the measures of the two red angles add up to the measure of the blue angle The second statement says that the two green angles add up to 180 An axiom about logic Axiom 9 Equals can be substituted for equals Two axioms about parallel lines Axiom 10 Given a point P and a line Z there is exactly one line through P parallel to Z Axiom 11 If and Z are parallel lines and m is a line that meets them both then alternate interior angles are equal That is Axiom 11 says that the two red angles are equal in the following picture provided that the horizontal lines are parallel Axiom 10 which is called Playfairls Axiom and Axiom ll distinguish Euclidean geometry from other geometries such as spherical geometry which we7ve talked a little about and hyperbolic geometry which well see eventuallyi Both axioms are intuitively correct but it took mathematicians a long time to realize that it was possible to do geometry without themi Now for two axioms that connect number and geometry Axiom 12 For any positive whole number n and distinct points A B there is some C between A B such thatnACAB Axiom 13 For any positive whole number n and angle AABC there is a point D between A and C such that n mAABD mAABC 4 Some theorems Now that we have a bunch of axioms and de nitions in place we can start using them to prove theoremsi We reserve the right to add more axioms and de nitions later if we need to Many of these theorems may seem obvious but that7s the point even seemingly obvious statements need to be proved The rst theorem was actually one of Euclidls original ve postulates axiomsi In our axiom system which is not the same as Euclid7s we don7t need to make it an axiomiwe can prove it from the axioms and de nitions abovei Theorem 1 All right angles have the same measure namely 90 Proof Suppose that AABX is a right angle By De nition 3 this means that the segments E and H can be extended to perpendicular lines ill Bgt and Let C be a point on ill Bgt such that B is between A and C Now by De nition 3 of perpendicular we know that mZABX mAXBC and by Axiom 8 we know that mAABXerAXBC 180 3 Substituting the rst equation into the second we nd that ZmAABC 180 so mAABC 900 D Notice that this proof says explicitly when it using an axiom or de nition This is an important habit to acquire when writing proofsiitls just like citing your sources By the way the little box at the end is a sign that the proof is complete The old practice was to use the abbreviation QiEiDi a Latin acronym meaning i i iwhich was to be proven The next several theorems say that certain things are unique every line segment has exactly one midpoint every angle has exactly one bisector and every line has exactly one perpendicular through a point on it De nition 7 A midpoint of a line segment E is a point C on E such that AC BC and 2 AC AB Theorem 2 Every line segment E has exactly one midpoint Proof First we show that E has at least one midpoint By Axiom 12 we can nd a point C between A and B that is on E such that 2 AC AB 1 So we just need to show that AC BC By Axiom 4 we also know that AC BC AB so substituting for AB in equation 1 which we can do by Axiom 9 gives us 2 AC AC BC and subtracting AC from both sides gives AC BC The second part of the proof is to show that E has only one midpoint To do this suppose that C and D are both midpointsithe goal is then to show that in fact C D By Axiom 3 we can do this by showing that CD 0 We don7t know anything about CD directly what we do know is that since C and D are both midpoints of E ACCBADDB 2 Now C is either between A and D or between B and D If C is between A and D then Axiom 4 says that AC CD AD But since AC AD by equation 2 this means that CD 0 If C is between B and D then Axiom 4 says that BC CD BD But since BC BD by equation 2 this means that CD 0 In either case Axiom 3 tells us that C D 1 Notice that the two cases were essentially the same In a written proof you might the author say something like Without loss of generality we7ll just consider the rst casellithatls what this means De nition 8 A bisector of an angle ABAC is a line A D such that D is between B and C and mABAD mADAC mABAC Theorem 3 Every angle ABAC has exactly one bisector Proof By Axiom 13 ABAC has at least one bisector We have to show that it has only one Suppose that A D and ill E are both bisectors Then D and E are points on W So D is either between B and E or between E and C Without loss of generality we7ll consider the rst case By Axiom 8 mABAD mADAE mABAE gt gt On the other hand since AD and AE are both bisectors we know from De nition 8 that mABAD mABAE gmABAC Substituting this into the previous equation gives mABAD mADAE mABAD which implies that mADAE 0 By Axiom 6 the points ADE are collinear 7 but that means that A D ill E so both bisectors are the same 1 As an immediate corollary we get the following uniqueness result Theorem 4 IfC is between A and B then there is exactly one line Z passing through C that is perpendicular to E Proof Suppose that Z is such a line Then Z is a bisector of AACBl By Theorem 3 there is exactly one such ne 1 Before we start looking at congruence and similarity we need to establish a few more theoremsl Theorem 5 Any two distinct lines intersect in at most one point Proof Let m and n be lines that have two different points P Q in common By Axiom 1 there is exactly one line containing P and Both m and n must be that line Therefore m n D Just like Axiom 1 this is a statement that seems utterly obvious but fails in spherical geometry where every pair of distinct lines meets in two polar opposite pointsl Theorem 6 The sum of the interior angles of any triangle is 180 That is if AABC is any triangle then mAABC szAC mAACB 1800 Proof Draw a line Z through C parallel to By Axiom 10 there is exactly one such liner Put points D and E on E so that C is between D and El By Axiom 8 and Axiom 6 ozquot y of 7 5 mAECB m1BCD mAECD 1800l By Axiom 11 a 1 and Substituting into the last equation gives a y B 1800 D Theorem 7 Suppose that two distinct lines mm both intersect a third line n If alternate interior angles are equal or corresponding angles are equal then m and m are parallel Proof Here s the picture Let7s prove the alternate interior angles77 case We are given that a 0 and we want to prove that m and m are parallel Suppose that m and 721 meet at a point Zi Then we have a triangle AAZZi By Theorem 6 mAAZZ m4AZ Z mAZAZ a B mAZAZ 1809 On the other hand a 5 180 by Axiom 8 Therefore mAZAZ 0 which says that ZAZ are collinearibut they7re not This is a contradiction and we conclude that m and 721 do not meet As for the corresponding angles77 case alternate interior angles such as a and 0 are equal if and only if corresponding angles such as a and a are equali This is because of the Vertical Angle Theorem which says that 0 a 1 5 Congruence and similarity De nition 9 Two things are congruent iff one of them can be moved rigidly so that it coincides with the other In particular if one of them consists of line segments then so does the other and corresponding sides have the same measurei We write 7 E Q to mean that f and Q are congruenti De nition 10 Two things are similar iff one of them is proportional to the other In particular if one of them consists of line segments then so does the other and corresponding sides have proportional measuresi We write 7 N Q to mean that f and Q are similar Notice that congruent77 is a stronger relationship than similar If two things are congruent then they are necessarily similar but two similar things don7t have to be congruenti Axiom 14 SSS Two triangles are congruent i their corresponding sides are equal That is AABC and AA B C are two triangles such that AB A B AC A C and BC B C then AABC E AA B C We already know from De nition 9 that the triangles are congruent then corresponding sides are equal What is new in Axiom 14 is the reverse implication corresponding sides are equal then the triangles are congruenti Axiom 15 AAA Two triangles are similar their corresponding angles are equal That is ifmlBAC mAB A C mAABC mAA B C and mABCA mAB C A then AABC N AA B C The abbreviations SSS and AAA are short for SideSideSide77 and AngleAngleAnglelli It is natural to ask about other criteria for congruence of triangles Theorem 8 ASA Two triangles are congruent i two pairs of corresponding angles and the sides between them are equal That is AABC E AABC i mABAC m4B A C mAABC m4A B C and AB ABl Proof To prove a theorem with an iff in its statement we need to prove that both implications holdl In this case it is easy to prove that if the triangles are congruent then the three equalities actually hold 7 this is an immediate consequence of De nition 9 So suppose we have two triangles that satisfy the three equalitiesl Draw then so that the points A B A B lie along the same line n in that order Also let a mABAC m1ABC 0 m4B A C mAABCl gt gt By Theorem 7 we know that AC and A C are parallel because they both intersect n and the corresponding 5 5 angles 10 are equal By s1milar logic we know that BC and BC are parallell Now slide AA B C along it so that the segments A B and E coincide with each other ie A A and B Bl We know we can do this because AB A B by hypothesis Theorem 7 still applies but here gt gt it says that AC and A C are actually the same line they canlt be parallel because they both contain the point A A so according to theorem the only other possibility is that they weren t distinct lines to begin gt with Similarly R and B C coincidel gt gt gt Let Z AC A C and m BC BC l The lines Z and m intersect in a unique point by Theorem 5 But that unique point must be both C and C 7 so we conclude that C C and the proof is done D An equivalent way of stating the ASA theorem is as follows A triangle is determined up to congruence by one side and the two angles adjacent to it In other words if you know the length of E and the angles a and B in the gure above then there7s only one possible point where C can be hence only one possible triangle AABCi Here s another congruence theoremi Theorem 9 SAS Two triangles are congruent two pairs of corresponding sides and the angles between those sides are equal That is AABC E AA B C i AB A B AC ACquot and mBAC mB A C i Proof Again if the triangles are congruent then the three equalities do indeed hold by De nition 9 As in the proof of Theorem 8 before we draw the triangles so that A BA B are collinear and we can gt 5 gt conclude from Theorem 7 that AC and AC are parallel but not BC and B C since we don7t know whether gt or not 6 equals Once again slide AA B C along n so that A A and B B so that now A A C i Now either C is between A and C or C is between A and Ch Without loss of generality suppose the rst case Then Axiom 4 says that AC C C ACi Also A and A are the same point so AC A C and substituting into the previous equation we get A C C C ACi But A C AC by hypothesis so C C 0 Therefore C C by Axiom 3 D An equivalent way of stating the SAS theorem is as follows A triangle is determined up to congruence by two sides and the angle between them It is important that the angle has to be between the sides that s why we call it SAS and not SSAi Specifying two sides and an angle opposite one of the sides does not determine the triangle up to congruence 7 see problem SA 20 Here are some important consequence of the angle congruence theoremsi Theorem 10 The base angles of an iosceles triangle are equal That is ifAB AC then ABC ACB First proof Let D be the midpoint of mi Then AD AD AB AC given and BD CD by de nition of midpointi So AABD E AACD by SSSi By de nition of congruence ABC 2 ACBi 1 Second proof Let E be the bisector of angle BAC and let E be the point where Z meets mi Then AE AE AB AC given and BAE E CAE by de nition of angle bisector So AABD E AACD by SAS Again by de nition of congruence ABC 2 ACBi D Third and slickest proof Observe that AB AC AC AB and BC CBi Therefore AABC ABAC by SSSi In particular ABC 2 ACBi The big idea here is that the triangle is congruent to a re ected copy of itself 1 B D C B E C B C Proof 1 Proof 2 Proof 3 The points D and E are actually the same point but we can t assume that from the start 7 so it is a logical mistake to say something like Let E be the bisector of ABAC and let Z meet W at its midpoint h To put it another way you can t assume that you can construct Z in a way that meets both those specifications On the other hand we can now prove that D and E coincide Finally two more very useful theorems about triangles inscribed in a semicircler Theorem 11 Suppose that E is a diamater of a circle centered at O and that C is a point on the circle Then mAACB 90 3a and szOC 2szAo 3b Proof The segments 0A OB 0C are all radii of the circle so the triangles AOAC and AOBC are isoscelesr Therefore by Theorem 10 mAOAC mlOCA and mlOCB mlOBCr Leta mAOAC mlOCA and mlOCB mlOBCr By Axiom 8 and Theorem 6 2a 2B mAOAC szCAJr szCB szBC mABAC mAACB mAABC 180 from which it follows that a B 90 proving 3a 10 Now let 9 mlBOCi Then 9 2B 180 and we already know that 2a 2B 180 and combining these two equations yields 9 2a which proves 3b D 6 The Pythagorean Theorem Theorem 12 In a right triangle with legs of lengths a and b and hypotenuse of length 5 we have a2b2 02 Proof Suppose we are given such a right triangle AAWX where WX is the hypotenuse see gure left below Construct a square WXYZ using the hypotenuse as one side see gure right below 2 y W W a a A b X A b X Extend AW ngg ments E and E respectively so that AABY and AADZ are right angles and then extend BY and DZ until they meet at C See gure left below 11 a c c b 0 Y 0 Y W W o C c C b a a A b X B A b X a B By Theorem 6 mAAWX mAAXW mAWAX 180 4a and WAX is a right angle so mAAWX mAAXW 900 4b On the other hand by Axiom 8 mAAXWmAWXYmABXY 180 4c and AWXY is a right angle so mAAXWerABXY 900 4d Comparing 4b and 4d tells us that mAAWX mABXYi Repeating these arguments we nd that AWX E ABXY E ACYZ E ADZW and AXW ABYX CZY ADWZi Also WX XY YZ ZW by construction so by ASA AAWX E ABXY E ACYZ E ADZW and so AWBXCYDZa AXBYCZDWIL See gure ght above Now we calculate the area of the square WXYZ two ways On the one hand areaWXYZ 02 5a On the other hand areaWXYZ areaABCD 7 areaAAWX 7 areaABXY 7 areaACYZ 7 areaADZW a b2 7 4ab2 a2 2a 122 7 2a a2 122 5b Now equating 5a and 5b gives the Pythagorean Theoremi D 12 7 De nitions and theorems about quadrilaterals De nition 11 A quadrilateral Q ABCD consists of four points ABCD and the line segments EWmi The diagonals of Q are the line segments E and BDi The quadrilateral is called convex if the diagonals cross each other but E does not meet W and W does not meet mi All quadrilaterals we7ll consider will be convexi Theorem 13 EC 25 The angles of every quadrilateral add up to 360 Proof Draw the quadrilateral AB CD and the diagonal El Label the angles as shown A Then mAABC m1BCD mACDA mADAB B 7 9 6 5 a by Axiom 8 a 7956 180 180 360 by Theorem 6 1 gt gt gt De nition 12 A quadrilateral Q ABCD is a parallelogram if AB is parallel to CD and BC is parallel to It is a rectangle if AABCABCDACDAADAB are all right angles It is a rhombus if AB BC CD DAi It is a square if it is both a rectangle and a rhombus The next several theorems are about parallelogramsi Theorem 14 EC 27 In a parallelogram PQRS opposite sides and opposite angles are equal That is gt gt gt gt PQ is parallel to RS and P5 is parallel to QR then PQ RS and PS RQ 6a and PQR 2 RSP and QRS 2 ASPQi 6b Proof Draw the diagonal mi By Axiom ll SRP ARPQ and ASPR AQRPi Also PR RP Axiom 2 so by ASA Theorem 8 we have congruent triangles APRS E ARPQi In particular RS PQ and PS RQ QR proving 6a Also ESP 2 APQR which is one of the assertions of 6b and we can obtain the other equality by constructing the diagonal QS and arguing similarlyi El 13 The next theorem is a wellknown fact about parallelograms but does take a little effort to prove Once we know it it will be very useful in the following two results about special kinds of parallelograms that is rhombi and rectanglesi Theorem715 Lhe diagonals of every parallelogram bisect each other That is if PQRS is any parallelogram and X PR QS is the point where its diagonals meet then PX RX and QX 5X Proof By Axiom 11 and the fact that P XR are collinear we have w PR5 E ARPQ and similarly by Axiom 11 and the fact that Q X S are collinear AQSR E ASQP Moreover Theorem 14 tells us that PQ SRi Together with the underlined angle equalities and ASA we conclude that AXSR E AXQP 7 from which it follows that PX RX and QX SXi 1 Theorem 16 EG 28 The diagonals of parallelogram PQRS meet at a right angle if and only if the parallelogram is a rhombus Proof Part I Suppose that the diagonals W meet at a right angle Then PX PX QX SX and APXS APXQ the second equality by Theorem 15 and the third by Theorem 1 So APXS APXQ by SAS and in particular PS PQi By the same argument PQ QR RS SPi That is the parallelogram is a rhombusi Part II Suppose that the parallelogram is a rhombusi Then the triangles APXQ ARXQ ARXS APXS 8 are mutually congruent by SSSi In particular APXQ ARXQ ARXS APXSi But these four angles add up to 360 by Axiom 7 so each of the four must equal 900 D Theorem 17 EG 29 The diagonals of a parallelogram are congruent to each other and only the parallelogram is a rectangle 14 Proof Part I Suppose that PR QSi Then PX QX RX SX by Theorem 15 So each of the triangles APXQ ARXQ ARXS APXS is isosceles so AXPQ E AXQP AXRQ E AXQR AXES E AXSR AXPS E AXSPi The argument of Theorem 15 says that AXSR E AXQP see and likewise AXPS E AXRQ so AXES AXPQ and AXPS AXRQi Combining with the previous equalities we know that a mAXPQ mAXQP mAXRS mAXSR mAXRQ mAXQR mAXPS mAXSPi On the other hand adding up the eight angles just listed gives 360 by Theorem 13 Therefore a 6 90 and each angle of the quadrilateral is a B for example mAPQR mAPQX mAXQR a by Theorem 8 Therefore the parallelogram is a rectangle Part II Suppose that PQRS is a rectangle We could prove that PR QS by methods similar to the preVious results but there s a much easier way apply the Pythagorean Theorem which says that PQ2PS2 and PR RS2PS2 On the other hand PQ RS by Theorem 15 so QS PPM 1 15

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