Class Note for MATH 409 with Professor Martin at KU (2)
Class Note for MATH 409 with Professor Martin at KU (2)
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CHAPTER 3 NonNeutral Geometry This chapter s propositions differ from those of the previous one in that they depend on Postulate 5 for their validity Their proofs are therefore not valid in the context of non Euclidean geometry 1 Parallelism The first of the non neutral propositions is the converse of Proposition 2335 the last proposition of the previous chapter PROPOSITION 311I29 A straight line falling on parallel straight lines makes the alternate angles equal to one another the corresponding angles equal to one another and the interior angles on the same side equal to two right angles A B 1 3 2 C D H 4 F Figure 31 31 31 PARALLELISM GIVEN Straight lines AB II CD straight line EF intersecting AB and CD at G and H respectively Fig 31 TOPROVE 41 L2 L3 44 4243 2 rightangles PROOF By contradiction Suppose A 1 and A 2 are unequal Then it may be assumed without loss of generality that 41 gt L2 41 43 gt 42 43 but 4 1 43 2 right angles PN 2317 2 right angles gt 42 43 AB and CD intersect PT 5 This however contradicts the fact that AB II CD and so A 1 A 2 Moreover 41 43 42 44 2 right angles PN 2317 43 44 CN 3 and also A 2 43 2 right angles QED The following proposition provides an alternative and intuitively appealing characterization of parallel lines PROPOSITION 312 The locus of all points on one side of a straight line that are equidistant from it is a straight line See Exercise 11 32 31 PARALLELISM EXERCISES 31A 10 11 12 13 14 15 Prove that if two parallel straight lines are cut by a third line then the two bisectors of a pair of alternate interior angles are parallel to each other Prove that if a straight line is perpendicular to one of two parallel straight lines then it is also perpendicular to the other one Suppose AB J KL and CD J MN are all straight lines such that KL H MN Prove that AB H CD Suppose AB J KL and CD J MN are all straight lines such that KL Vl MN Prove that AB Vl CD Prove that two angles whose sides are respectively parallel are either equal or supplementary Prove that two angles whose sides are respectively perpendicular are either equal or supplementary In AABC AD is the bisector of A BAC and E is a point on AC such that DE H AB Prove that AE DE For a given A ABC AD H BC and AD AB Prove that BD bisects either the interior angle or the exterior angle at B Prove that if the points A B are on the same side of the straight line In and at the same distance from m then AB Hm Prove that if the points A B are such that AB H m then they are at the same distance from m Use the above two exercises to prove Proposition 312 Prove that the internal bisectors of each pair of angles of a triangle intersect Given two distinct parallel lines construct a straight line that is parallel to both and also equidistant from both Comment on Proposition 311 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 312 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi PROPOSITION 313I30 Distinct Straight lines parallel to the same straight line are also parallel to one another GIVEN Distinct straight lines AB EF CD EF Fig 32 T0 PROVE AB II CD 33 31 PARALLELISM A D I C 1 B J 3 E F Figure 32 PROOF By contradiction Suppose AB and CD intersect in some point I Join I to any point J of EF Then 41 43 PN311 AB IIEF 42 43 PN311 CDIIEF 41 A 2 CN 1 but this is impossible since the straight lines AB and CD are distinct Hence AB II CD QED Euclid begins his proof of this proposition by drawing a straight line PQ that intersects all the three given lines While intuitively plausible the existence of such a line calls for a justification and Euclid39s proof is therefore incomplete The need for such a justification is demonstrated by Figure 33 which exhibits three pairwise parallel hyperbolic geodesics such that no single geodesic intersects all three Ann Figure 33 Three hyperbolic parallel straight lines that are not intersected by the same hyperbolic straight line 34 31 PARALLELISM EXERCISES 3 IB 1 If a straight line intersects one of two parallel straight lines in only one point then it also intersects the other one 2 Comment on Proposition 313 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi PROPOSITION 314I31 Through a given point to draw a straight line parallel to a given straight line GIVEN Straight line BC point A not on FC Fig 34 TO CONSTRUCT A straight line AE such that AE BC Figure 34 CONSTRUCTION Let D be any point on BC and draw AD Construct A DAE 4 ADC PN 2328 Then AE is the required straight line PROOF A EAD 4 CDA Construction AE BC PN 2335 QED The following proposition has supplanted Euclid39s Postulate 5 in many texts where it is known as Playfair39s Postulate Although this will not be demonstrated here the two are in fact logically equivalent 35 31 PARALLELISM PROPOSITION 315Playfair s Postulate Through a point not on a given straight line there exists exactly one straight line that is parallel to the given line See Exercises 1 and 2 Just like Postulate 5 Playfair s postulate does not hold in hyperbolic geometry Figure 35 exhibits three distinct geodesics p q r all of which contain the same point P and all of which are parallel to the same geodesic m Figure 35 A hyperbolic counterexample to Playfair s Postulate It is now possible to give a more precise definition of hyperbolic geometry This calls for negating Playfair s postulate which is equivalent to Postulate 5 In View of Proposition 237 the following postulate is the proper negation of Playfair s postulate H Hyperbolic There exists a straight line that is parallel to two intersecting distinct straight lines Hyperbolic geometry is the geometry based on Euclid s Postulates 1 2 3 4 A S and Postulate H 36 31 PARALLELISM PROPOSITION 316032 In any triangle if one of the sides be produced the exterior angle is equal to the two interior and opposite angles and the three interior angles of the triangle are equal to two right angles GIVEN AABC side BC extended to D Fig 36 TO PROVE 4ACD 4ABC 4 CAB 4ABC 4 BCA 4 CAB 2 right angles E D Figure36 PROOF Draw CE IIAB PN 314 Then 45 42 PN311 46 41 PN311 44 2 42 41 CN2 4443 414243 CN2 2 right angles 2 41 42 43 PN 2317 QED Recall that according to Chapter 1 the sum of the angles of every spherical triangle is greater than 1800 PN 115 whereas the sum of the angles of every hyperbolic triangle is less than 1800 PN 126 EXERCISES 31C 37 31 PARALLELISM 1 Prove Proposition 315 2 Prove that in AABC the bisector of the exterior angle at A is parallel to BC if and only if AB AC 3 Prove that a straight line that is parallel to one side of an isosceles triangle cuts off another isosceles triangle Note There are two distinct cases to be considered here 4 A straight line cuts off an isosceles triangle from a given isosceles triangle Prove that the straight line is parallel to one of the sides of the given isosceles triangle 5 In an isosceles AABC a line perpendicular to the base BC intersects AB and AC in the points D and E respectively Prove that AADE is also isosceles 6 In AABC 4 BAC 900 and 4 ACB 30 Prove that BC 2AB 7 In AABC 4 ABC 600 and BC 2AB Prove that AABC is a right triangle 8 Prove that in a right triangle the angle between the altitude to the hypotenuse and one of the legs equals the angle opposite that leg 9 Let D be that point on side BC of A ABC such that AD is the bisector of A BAC Prove that 4 ADC is half the surn of the interior angle at B and the eXterior angle at C 10 Prove that in AABC the bisectors of the interior angle at B and the exterior angle at A form an angle that is half the interior angle at C 11 Prove that in A ABC the angle bisector and the altitude at A form an angle that is half the difference between the interior angles at B and C 12 Prove that in a right A ABC the bisector of 4 ABC the altitude to the hypotenuse BC and the side AC form an isosceles triangle 13 The point D on the hypotenuse BC of the right isosceles A ABC is such that BD AB Prove that 4 BAD 6750 14 Prove that if the diagonals of quadrilateral ABCD are equal and the sides AB CD then AD H BC 15 Prove that the surn of the interior angles of a quadrilateral is 3600 16 Prove that in quadrilateral ABCD the bisectors of the interior angles at A and B form an angle that is half the surn of the interior angles at C and D and if the bisectors of the interior angles at A and C intersect they form an angle that is half the difference between the angles at Band D A polygon is said to be convex if all of its diagonals fall in its interior 17 Prove that the surn of the interior angles of a conveX nisided polygon is n 7 21800 18 Prove that the surn of the interior angles of an arbitrary nisided polygon is n 7 21800 Go ahead and use the difficult to prove fact that every polygon has a diagonal that lies completely 38 23 24 25 26 27 28 29 30 31 32 31 PARALLELISM inside it 19 Prove that the number of acute interior angles of a convex polygon cannot exceed 3 20 Prove that the sum of the exterior angles of a convex polygon is 3600 Try to prove this without making use of Exercise 17 above 21 Construct AABC given the data a A B 22 Construct an isosceles triangle given one of its angles and one of its sides Construct AABC given the data a b c A Construct AABC given the data b c A B Comment on Proposition 314 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 315 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 316 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Explain why there are no rectangles in spherical geometry Explain why there are no rectangles in hyperbolic geometry Are there rectangles in taxicab geometry Are there rectangles in maxi geometry The following method for trisecting an arbitrary angle is credited to Archimedes If that attribution is correct he must have been aware of its shortcomings as a construction in the sense of Euclid Let a be a given angle with vertex A Fig 37 Draw a circle of radius AB 2 AC On a ruler mark two points D and E such that DE 2 AB 2 AC and place the ruler on the page so that the point E falls on the extension of AB the point D falls on the circle A AB and the ruler also passes through the point C Prove the following assertions a LADCLACD7 b L AED L EAD y c a3y or y 13 Figure 37 An angle quottrisectionquot 39 33 34 35C 36C 31 PARALLELISM Explain why this quottrisectionquot of or does not meet Euclid39s standards for a construction Criticize the following quotneutral proofll of Playfair39s Postulate offered by Proclus 4107485 quotI say that if any straight line cuts one of two parallels it will cut the other also For let AB CD be parallel and let EFG cut AB at F with G between AB and CD I say that it will cut CD also For since BF FG are two straight lines from one point F they have when produced indefinitely a distance greater than any magnitude so that it will be greater than the interval between the parallels Whenever therefore they are at a distance from one another greater than the distance between the parallels FG will cut CDll Criticize the following quotproofquot of the fact that the surn of the interior angles of a triangle is 1800 Let ABC be a given triangle let d be a line segment that lies on the straight line AB with its center at A Slide d along AB until its center falls on B and then rotate it through the exterior of the triangle about B as a pivot until it falls along side BC Next slide d along BC until its center reaches C and rotate it about C as a pivot through the exterior of the triangle until it falls along CA Finally slide d along CA until its center reaches A and rotate it about A as pivot so that it comes into its initial position If the triangle39s interior angles are a 3 y then the segment d has been rotated successively by the angles 1800 7 1800 7 y and 1800 7 a before it returned to its original position Consequently 1800 7 8 1800 7 y 1800 7 a 3600 from which it follows that a 3 y 1800 Perform the construction of Proposition 314 using a computer application Use a computer application to verify Proposition 316 Euclid39s statement of the following proposition is awkward and so it appears here in a paraphrased forrn PROPOSITION 317I33 A quadrilateral in which two opposite sides are both equal and parallel to each other is a parallelogram GIVEN Straight line segments AB II CD AB 2 CD Fig 38 T0 PROVE AC M ED AC 2 BD 310 31 PARALLELISM B A 9 A D C Figure 38 PROOF Draw BD BC AC Then A ABC 2 A DCB by SAS because AB 2 DC Given 4 1 A 2 Alternating angles AB II CD PN 311 BC 2 CB AC 2 DB and A 4 A 3 AC BD Equal alternating angles PN 2334 QED PROPOSITION 318I34 If both pairs of opposite sides of a quadrilateral are parallel to one another then they as well as the opposite angles are equal to one another and the diameter bisects the area GIVEN Quadrilateral ACDB AB II CD AC BD TO PROVE AB CD AC BD 4 CAB 4 BDC 4 ABD 4 DCA A ABC A DCB ABDC 311 PROOF Also 31 PARALLELISM I v A Figure 39 A ABC 5 A DCB by ASA because A 5 A 6 Alternating angles AB II CD PN 311 BC 2 CB 4 7 A 8 Alternating angles AC DB PN 311 AB 2 CD AC 2 BD 43 44 A ABC 2 A DCB 39ABDC A 1 A 2 CN 2 QED EXERCISES 31D SQMPW Both pairs of opposite sides of a quadrilateral are equal to each other Prove that the quadrilateral is a parallelogram Both pairs of opposite angles of a quadrilateral are equal to each other Prove that the quadrilateral is a parallelogram Prove that the diagonals of a parallelogram bisect each other Prove that if the diagonals of a quadrilateral bisect each other then it is a parallelogram Prove that a parallelogram is a rectangle if and only if its diagonals are equal to each other Prove that a parallelogram is a rhombus if and only if its diagonals are perpendicular to each other Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and equals half its length 312 10 11 12 13 14 15 16 17 18 19 20 21 22 23 31 PARALLELISM The midpoint of side AB of AABC is D and E is a point of AC such that DE H BC Prove that AE EC and DE BC2 Prove that the midpoints of the four sides of a quadrilateral are the vertices of a parallelogram Prove that each of two medians of a triangle is divided by their intersections into two segments one of which is double the other Prove that the three medians of a triangle all pass through one point Point E in the interior of square ABCD is such that A ABE A BAE 150 Prove that A CDE is equilateral In AABC AB 2 AC and D E F are points on the interiors of sides BC AB AC respectively such that DE J AB and DF J AC Prove that the value of DE DF is independent of the location of D Prove that the three segments joining the midpoints of the three sides of a triangle divide it into four congruent triangles Prove that three parallel straight lines that cut off equal line segments on one straight lines also cut off equal line segments on every straight line that intersects them Hint Through the middle intersection point on one straight line draw a line parallel to the other straight line A trapezoid is a quadrilateral two of whose sides are parallel Prove that the line segment joining the midpoints of the noniparallel sides of a trapezoid is parallel to the other two sides and equals half the sum of their lengths Construct angles of the following magnitudes a 60 b 30 c 1200 d 75 Through a given point construct a straight line such that its portion between two given parallel straight lines is equal to a given line segment Let A be a point in the interior of an angle Construct a straight line whose segment between the sides of the angle has A as its midpoint A pair of parallel straight lines is intersected by another pair of parallel straight lines Through a given point constr39uct another straight line on which the two given pairs cut off equal line segments Given an angle determine the locus of all the points the sums of whose distances from the sides of the angle equals a given magnitude In a given AABC constr39uct points M on AB and N on BC such that BM NC MN and VIN H BC Construct AABC given the data a 1 ha I3 b 1 ha hb c a hb a d hb he a e hb me a f a he b c g abc y h abc ha 313 24 25 26 27 28 29 30 31 PARALLELISM Construct a parallelogram given a two adjacent sides and the included angle b two adjacent sides and a diagonal c two adjacent sides and the distance between two opposite sides d a side and the two diagonals e the diagonals and the angles between them Construct a rectangle given one side and the diagonal Construct a rhombus given a its side and one of its angles b its side and one diagonal c both diagonals Construct a square given a its side b its diagonals Construct AABC given the data a womb b ahamb Comment on Proposition 317 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Comment on Proposition 318 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 2 Area Euclid defined the concept of area by means of axioms that he called Common Notions This axiomatic approach is customary today as well although the specific axioms are different from those used by Euclid The modern approach to area stipulates that a certain unit of length called unit has been chosen The square the length of whose side is 314 32 AREA Figure 310 1 unit Fig 310 is denoted by unit square or unit2 and serves as the unit for measuring areas It is then assumed that area is a measurement of figures that satisfies the three properties or axioms listed below UNIT The unit square has area 1 unit2 ADDITIVITY If a figure is divided by a line into two subfigures then the area of the figure equals the sum of the areas of the subfigures INVARIAN CE Congruent figures have equal areas Loosely speaking the additivity and invariance axioms were stated by Euclid as Common Notions 2 and 4 respectively The unit axiom however has no analog in Euclid39s system As a consequence Euclid39s Elements contains no proposition that computes areas explicitly Instead Euclid made comparative statements such as parallelograms on equal bases and between the same parallels are equal and if a parallelogram have the same base with a triangle and be in the same parallels the parallelogram is double the triangle This has the theoretical advantage of dispensing with units and the practical disadvantage of not answering the reasonable question of what is the area of a rectangle of dimensions 315 32 AREA 3 and 5 stadia Greek mathematicians did of course make use of units and could resolve such questions with ease It is just that Euclid for reasons that can only be guessed at and that in the author s opinion were probably esthetic decided to develop his geometry without any units whatsoever Propositions 321 and 322 below are the explicit modern day analogs of Euclid39s 135 PN 323 They give explicit formulas for the areas of rectangles and parallelograms Their complete proofs unfortunately contain elements that are beyond the scope of this text Specifically one runs into the difficulty inherent in proving propositions regarding line segments with irrational non fractional lengths These difficulties were first encountered by the Greeks in the sixth century BC and eventually surmounted by Eudoxus two hundred years later Euclid39s book did incorporate Eudoxus39s treatment of irrational numbers but it would be impractical to expound this theory here Instead a mere supporting argument for the fact that the area of a rectangle is given by the product of the lengths of its sides is offered It is customary in today39s high school geometry textbooks to circumvent these difficulties by stating this formula as yet another axiom the Rectangle Axiom In the author s opinion this is a misguided solution to a pedagogical problem since it opens up the possibility of stating many other interesting and non trivial geometrical facts as axioms even when elementary and convincing albeit logically incomplete arguments are available PROPOSITION 321 Ifa rectangle has dimensions a units and 17 units then it has area ab unitZ GIVEN ABCD with sides a 17 TO PROVE Area of ABCD 2 ab unit2 SUPPORTING ARGUMENT If a and b are positive integers then a rectangle of dimensions a and b can be divided into ab unit squares by means of straight lines that 316 32 AREA are parallel to its sides Fig 311 Consequently by the Additivity Property the given 2 rectangle has area ab un1t a3 Areaunitlunit2 l Area ofrectangle 15 unit2 Figure 311 Similarly if a rectangle has dimensions a 1m and b 1n for some positive integers m and n then the unit square can be divided into mn copies of the given rectangle all of which by the Invariance Property have the same area Fig 312 Hence the given rectangle has area 2 1 unit 1 mn m 2 2 unit 2 db un1t H 13 14 Area of rectangle 112 unit2 lunit2 Figure 312 317 32 AREA Next if a rectangle has dimensions a mn and b pq where m n p q are all positive integers then it can be decomposed into mp rectangles each of which has l l d1mens1ons n and F1g 313 Since each of these latter rectangles is now known to ml unit2 it follows from the Additivity Property that the given rectangle has have area area 2 2 unit 2 ab un1t E gr 1 2 mp unit 2 74 14 56 16 Area 124 unitz Area of rectangle 3524 Figure 313 This verifies the proposition for all rectangles with fractional dimensions As was mentioned above the extension of this formula to rectangles with arbitrary real dimensions lies beyond the scope of this text QED An altitude of a parallelogram is any line segment cut off by two opposite sides from a straight line that is perpendicular to both of them It follows from Proposition 312 that all the altitudes joining the same pair of opposite sides of a parallelogram have equal length PROPOSITION 322 The area of a parallelogram with base 17 units and altitude h units is bh unitZ 318 32 AREA GIVEN ABCD with base 17 and altitude h Fig 314 TO PROVE Area of 39ABCD bh unit2 Figure 314 PROOF In the given parallelogram draw AE 39 CD and BF 39CD with E F on CD Thus ABFE is a rectangle with area bh unit2 Since A ADE s A BCF it follows that they have the same area and hence by the Additivity Property ABCD ABFE bh unit2 QED Euclid39s version of Propositions 321 and 322 is now stated together with his proof as well as another proof that is more consistent with modern pedagogy PROPOSITION 323I35 Parallelograms which are on the same base and in the same parallels are equal to one another GIVEN 39ABCD and 39 EBCF such that A D E F are collinear Fig 315 TO PROVE 39ABCD 2 EBCF 319 32 AREA Figure 315 PROOF Euclid Since AD 2 BC 2 EF PN 34 it follows from CN 2 that AE DF Then ABAE s A CDF by SAS because See above AE DF 4 1 A 2 Corresp angles AB DC PN 311 AB 2 DC Parallelogram ABCD PN 318 A EAB A FDC ABGD EGCF Subtract A DGE CN 3 39ABCD 2 39EBCF Add A GBC CN 2 QED Euclid39s proof of Proposition 323 is incomplete albeit easily fixed because it depends on the relative position of the points A D E F on their common line Exercise 12 PROOF modern B J C Figure 316 Draw HJ perpendicular to AD and BC Fig 316 It then follows from PN 322 that NABCD BCHJ NEBCF QED 320 32 AREA PROPOSITION 324036 Parallelograms which are on equal bases and in the same parallels are equal to one another See Exercise 1 The area of the triangle will be given the same dual treatment as that of the rectangle First the modern formula is offered PROPOSITION 325 The area of a triangle with base 17 units and altitude h unitsis bhZ unitZ A b C Figure 317 PROOF Through the vertices B and C of AABC draw straight lines parallel to AC and AB respectively and let their intersection be D Fig 317 It is clear that ACDB is a parallelogram and hence by Proposition 318 bh AABC ACDB 7 unit2 NIH QED Next comes Euclid39s version 321 32 AREA PROPOSITION 326037 Triangles which are on the same base and in the same parallels are equal to one another GIVEN AABC A DBC AD II BC Fig 318 TO PROVE A ABC 2 A DBC A D E F B C Figure 318 PROOF Let E be the intersection of AD with the straight line through B parallel to AC and let F be the intersection of AD with the straight line through C parallel to BD PN 314 Then AEBC 39DBCF PN 322 A ABC 2 39AEBC PN 318 A DEC 2 DBCF PN 318 A ABC 2 A DBC QED PROPOSITION 327038 Triangles which are on equal bases and in the same parallels are equal to one another See Exercise 2 PROPOSITION 328039 Equal triangles which are on the same base and on the same side are also in the same parallels 322 32 AREA GIVEN A ABC 2 A DBC A and D are on the same side of BC TO PROVE AD BC Figure 3 19 PROOF By contradiction Suppose AD and BC are not parallel and let E be the intersection of BD with the straight line through A parallel to BC Then A ABC 2 A EBC PN 326 A EBC lt A DBC CN 5 AABC lt A DBC This however contradicts the give equality of the two triangles Hence AD BC QED PROPOSITION 329I40 Equal triangles which are on equal bases and on the same side are also in the same parallels According to Heath Proposition 329 is an interpolation into The Elements by a later georneter Exercise 2 PROPOSITION 3210I41 If a parallelogram have the same base with a triangle and be in the same parallels the parallelogram is double of the triangle See Exercise 2 323 32 AREA There is no analog of Proposition 325 for the area of a general quadrilateral In practice any such quadrilateral can be divided into triangles by means of a diagonal and then the area of each of the parts can be evaluated by means of Proposition 325 A similar procedure can be used to dissect any polygon regardless of the number of its sides into triangles Neither Euclid39s nor the modern approach to areas are applicable to spherical geometry Both of these approaches rely heavily on the notion of parallelism and the sphere has no parallel geodesics Thus another approach is required in order to develop a theory of spherical areas As spherical polygons can also be dissected into spherical triangles it suffices to provide a formula for the latter It is clear that any two lunes of the same angle a on the same sphere can be made congruent by a series of rotations of that sphere Consequently every two such lunes have the same area This in turn implies that the area of a lune is proportional to its angle Since the lune of angle 2n radians has area 471R2 the sphere s total surface area the following lemma is obtained LEMMA 3211 On a sphere of radius R the area of a lune of angle a radians is 2 2 20cR unit The following theorem was first discovered by the Flemish mathematician Albert Girard 1595 1632 The proof presented here is due to Euler PROPOSITION 3212 On a sphere of radius R the area of the spherical triangle ABC with angles of radian measures a it y is a it y nR2 unit2 GIVEN Spherical A ABC with interior angles a y measured in radians TO PROVE A ABC a f y nR2 unitZ 324 32 AREA Figure 320 PROOF Let A B C be the respective antipodes of A B C Fig 320 Draw the great circles that contain the geodesic segments AB BC and CA The hemisphere in front of the great circle BCB C is thereby divided into four spherical triangles ABC AB C AB C ABC whose areas are denoted respectively by T1 T2 T3 T4 From the construction it follows that the spherical A ABC is congruent to the A AB C of area T2 Hence T1 T2 lunea Similarly T1T3 lune and 325 32 AREA T1 T4 lune y Consequently 2T1 lunealune luney T1T2T3T4 2a 23 2 2nR2 unit2 and the statement of the theorem now follows immediately QED EXERCISES 32A 1 Prove Proposition 324 2 Use Proposition 325 to prove a Proposition 326 b Proposition 327 0 Proposition 328 d Proposition 329 e Proposition 3210 3 One of the triangle39s sides is divided into 11 equal segments and the division points are joined to the opposite vertex Prove that the triangle is divided into 11 equal parts 4 Prove that the area of the trapezoid equals the product of half the sum of its parallel sides with the distance between them 5 Prove that the diagonals of a parallelogram divide it into four equal triangles 6 Prove that the line segmentjoining the midpoints of two sides of a triangle cuts off a triangle that is equal to one fourth of the original triangle 7 Prove that the parallelogram formed by the midpoints of the sides of a quadrilateral equals one half of that quadrilateral 8 Prove that the triangle s medians divide it into siX equal triangles 9 The diagonals of a quadrilateral divide it into four equal triangles Prove that the quadrilateral is a parallelogram 326 32 AREA 10 Prove that if the point P lies in the interior of ABCD then the parallelogram equals twice the sum of AABP and A CDP 11 Each of the sides AB BC CA of an equilateral triangle is extended by their common length to points D E F respectively all in the same sense Prove that A DEF 7A ABC 12 Complete Euclid39s proof of Proposition 323 Both the taxicab and maxi areas of a gure are de ned to equal its Euclidean area 13 Comment on Propositions 321 322 and 325 in the context of taxicab geometry 14 Comment on Proposition 323 324 and 32610 in the context of taxicab geometry 15 Comment on Propositions 321 322 and 325 in the context of maxi geometry 16 Comment on Proposition 323 324 and 32610 in the context of maxi geometry Euclid s Propositions 142 45 are of limited interest They are included here only in order to facilitate the later discussion of the Golden Ratio Proposition 341 PROPOSITION 3213I42 To construct in a given rectilineal angle a parallelogram equal to a given triangle See Exercise 1 The above proposition is an example of a conversion which consists of the construction of a polygon IT of some prespecified nature that is equal to a given polygon H PROPOSITION 3214I43 In any parallelogram the complements of the parallelograms about the diagonal are equal to one another GIVEN 39ABCD K is a point on the diagonal AC BGKE 39 KFDH Fig 321 TO PROVE 39BGKE KFDH 327 32 AREA A H D E V B G C Figure 321 PROOF See Exercise 2 PROPOSITION 3215I44 To a given straight line to apply in a given rectilineal angle a parallelogram equal to a given triangle See Exercise 3 PROPOSITION 3216I45 To construct in a given rectilineal angle a parallelogram equal to a given rectilineal figure See Exercise 4 EXERCISES 32B H Prove Proposition 3213 Prove Proposition 3214 Prove Proposition 3215 Prove Proposition 3216 Convert a given parallelogram into a rectangle with the same base Convert a given parallelogram into a rhombus with the same base Convert a given parallelogram into another parallelogram with the same base and a given angle Convert a given parallelogram into a triangle with the same base and a given angle WPO QMPWN Convert a given triangle into a right triangle with the same base H 0 Convert a given triangle into an isosceles triangle with the same base H H Convert a given triangle into another Uiangle with the same base and a given angle 328 32 AREA 12 Bisect the area of a parallelogram by means of a straight line that is parallel to a given straight line 13 Given AABC construct apoint O in its interior such that the triangles AOB BOC COA all have equal areas 3 The Theorem of Pythagoras The Theorem of Pythagoras was discovered independently by several cultures and has been given more different proofs than any other theorem It is considered by many mathematicians to be the most important of all theorems and has the dubious distinction of being misquoted in the classic movie The Wizard of Oz and of being the subject of popular jokes It will be presented following an easy lemma PROPOSITION 331I46 On a given straight line to describe a square GIVEN Line segment AB Fig 322 TO CONSTRUCT quot ABCD E C D J2 4K 1 3 f A E Figure 322 CONSTRUCTION Draw EA JAB PN 2311 and let D on AE be such that AD 2 AB Let C be the intersection of the straight lines through B and D that are parallel 329 33 THE THEOREM OF PYTHAGORAS to AD and AB respectively PN 314 Then quadrilateral ABCD is the required square PROOF By construction ABCD is a parallelogram Since AB 2 AD it follows that AB 2 AD 2 DC 2 CB It remains to show that all of the angles of ABCD are right angles However 41 A 2 2 right angles PN 311 A 2 right angle A 1 is a right angle A 3 A 4 right angle PN 318 QED PROPOSITION 332I47 The Theorem of Pythagoras In rightangled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle GIVEN AABC A BAC right angle quotABFG quotACKH 39quot BCED Fig 323 TO PROVE quotBCED quot39ABFG quotACKH H G K A F B J M C D L E Figure 323 330 33 THE THEOREM OF PYTHAGORAS PROOF Let L M be the respective intersections of the straight lines DE and BC with the straight line through A parallel to BD and CE PN 314 Note that the points G A C are collinear A GAB LBAC 2 right angles PN 2318 the points B A H are collinear A HAC LBAC 2 right angles PN 2318 A ABD s A FBC by SAS because BD 2 BC Sides of the same square 4 ABD A FBC Both equal 4 ABC right angle AB 2 FB Sides of the same square A ABD A F BC BDLM 39quot ABF G Doubles of equal triangles PN 3210 A similar argument yields the equation CELM quot ACKH and hence TBCED TBDLM fCELM EABFG TACKH QED Two other proofs of this theorem are now sketched out If a and b are the legs and c is the hypotenuse of a right triangle then the square of side a b can be dissected in the two ways depicted in Figure 324 The dissection of I calls for no explication That of 11 requires a proof that the interior quadrilateral labeled as 02 is indeed a square see Exercise 8 However once these dissections are granted it is clear from Figure 324 that a2 b2 02 This proof is attributed to the Chinese mathematician Chou pei Suan ching who lived circa 250 BC 331 33 THE THEOREM OF PYTHAGORAS Figure 324 The next proof is due to the Indian mathematician Bhaskara 1114 1185 The square of side 0 can be dissected in the manner depicted in Figure 325 It then follows that 2 2 2 c 42ab 2aba2abb ab ab2 Figure 325 Yet another proof of the Theorem of Pythagoras in indicated in Exercise 17 This one is due to president James Garfield 183 1 1881 The Theorem of Pythagoras has a converse 332 33 THE THEOREM OF PYTHAGORAS PROPOSITION 333I48 If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle the angle contained by the remaining two sides is right See Exercise 9 2 2 2 Since 3 4 5 it follows that any tr1angle whose s1des have lengths 3 units 4 units and 5 units is necessarily a right triangle So is the triangle whose sides have lengths 5 12 13 Triples ofintegers a b c such that are known as Pythagorean triples but the interest in such triples precedes Pythagoras by over a thousand year The Babylonian tablet PLIMPTON 322 dated between 1900 and 1600 BC contains fifteen Pythagorean triples the largest of which consists of 12709 13500 and 18541 Although it is highly unlikely that the Babylonians found these numbers by trial and error it is not known what method they used to generate these triples Not surprisingly the earliest method for generating Pythagorean triples appears in Euclid39s The Elements Lemma 1 to Proposition 29 of Book X states that if m gt n are any positive integers then 2mn2 m2 n22 m2 n22 2 2 2 2 Exercise 13 so that 2mn m n m n form a Pythagorean tr1ple For example m 5 and n 4 yield the triple 2542 52 422 52422 or 333 33 THE THEOREM OF PYTHAGORAS 40292 41 Pierre Fermat 1601 1665 took it for granted that Euclid39s method can be used to generate all the Pythagorean triples and this fact was proven by Euler a hundred years later Specifically Euler proved that if a b c are numbers whose only common divisor is 1 and which constitute a Pythagorean triple then there exists a pair of relatively prime integers m n such that a b c 2 2mm m2 n2 m2 n2 All other Pythagorean triple of course proportional to these EXERCISES 33A 10 Which of the following triples of numbers are the lengths of the sides of a right triangle 5 12 13 57302 491714 650463 xE Show that an equilateral triangle of side a has area T a 7 10 15 b c 203750 364056 417194 d An isosceles right triangle has a hypotenuse of length 0 Compute its other sides and its area A right triangle has an angle of 300 and a hypotenuse of length 1 Compute its other sides and its area Compute the area of a rhombus whose sides equal 13 and one of whose diagonals has length 10 Compute the area of a parallelogram whose sides have lengths 11 and 8 and one of whose angles is 450 The diagonals and one side of a parallelogram have lengths 30 16 17 respectively Prove that it is a rhombus and compute its area Show that the interior quadrilateral in Dissection II of Figure 324 is indeed a square of area 02 Prove Proposition 333 Find the error in the following quotproofquot of the quotpropositionquot that every triangle is isosceles GIVEN AABC 334 11 12 13 14 15 16 17 33 THE THEOREM OF PYTHAGORAS TO PROVE AB AC M Figure 326 PROOF Let N be the intersection of the bisector of 4 ABC and the perpendicular bisector of side BC M the midpoint of BC and ND39AB NE AC Then ND NE PN 2333 AD 2 AE Pythagoras Also EN CN PN 2314 BD 2 CE Pythagoras AB AC CN 2 QED Given a square of side a construct a square of double its area Given a square of side a and a positive integer n construct a square Whose area equals 11 times that of the given square Construct a square Whose area equals the sum of three given squares Use algebra to prove that Euclid39s method does indeed generate Pythagorean triples Assume that a line segment of length 1 inch is given Prove that line segments of the following lengths can be constructed a 12 inch b 13 inch c d IE inch Where n is any positive integer e 5 inch WE inch Find two nonicongruent isosceles triangles Whose sides have integer lengths and Whose perimeters and areas are equal Prove the Theorem of Pythagoras by applying Exercise 32A4 to Figure 327 335 33 THE THEOREM OF PYTHAGORAS b a Figure 327 18C Perform the construction of Proposition 331 using a computer application 19C Use a computer application to verify the Theorem of Pythagoras Both spherical and hyperbolic geometry have their own versions of the Theorem of Pythagoras Their proofs follow directly from the appropriate trigonometries Exercises 1 3 PROPOSITION 334The spherical Theorem of Pythagoras If the spherical A ABC has a right angle at C then cos c cos a cos b PROPOSITION 335The hyperbolic Theorem of Pythagoras If the hyperbolic A ABC has a right angle at C then cosh c cosh a cosh b EXERCISES 33D l Derive the spherical Theorem of Pythagoras from Proposition 112 336 33 THE THEOREM OF PYTHAGORAS 2 Find the length of the hypotenuses of the three spherical isosceles right Triangles Whose legs have lengths l l 01 respectively Compare the answers to the lengths of the hypotenuses of the three Euclidean isosceles right triangles both of Whose legs have lengths l l 01 respectively 3 Derive the hyperbolic Theorem of Pythagoras frorn Propositions 122 4 Find the length of the hypotenuses of the three hyperbolic right triangles Whose legs have lengths l l 01 respectively Compare the answers to the lengths of the hypotenuses of the three Euclidean triangles both of Whose legs have lengths l l 01 respectively 5 Is there a taxicab version of the Theorem of Pythagoras 6 Is there a maxi version of the Theorem of Pythagoras 4 Consequences of the Theorem of Pythagoras optional Book II of Euclid39s Elements contains a variety of consequences of the Theorem of Pythagoras of which only a sample are presented here The first of these is tantamount to a construction of the Golden Ratio This proposition will be used later in the construction of the regular pentagon PROPOSITION 341H6 To cut a given line segment so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment GIVEN Line segment AB Fig 328 TO CONSTRUCT A point C on AB such that AB BC 2 AC2 D C Figure 328 337 34 CONSEQUENCES OF THE THEOREM OF PYTHAGORAS 1 CONSTRUCTION At B construct BD AB and BD 5 AB Join AD let E be the point on AD such that DE 2 DB and let C be the point on AB such that AC 2 AE Then C is the required point PROOF Set AB 2a Then DE 2 BD 2 a and AC AE AD DE 2a2a2 a 03 1a so that AC2 03 12a2 5 2x5 1a2 6 2x13a2 and ABBC ABABAC ABAB AE ABAB AD DE 2 2a2a IEa an 23 13a2 AC QED In the context of the above proposition the common value 7 of the ratios AC 2 AB AB E AC AD DE VB 1 IE 1 2a is called the Golden Ratio While of demonstrated mathematical interest this quantity has also been the subject of much nonsensical speculation Typical of this latter variety is an article that reports that the average ratio of the height of a man39s navel off the ground to his height equals the Golden ratio The following two propositions constitute Euclid39s analog of the modern day Law of Cosines Their proofs are relegated to the exercises PROPOSITION 342II12 In obtuseangled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse 338 34 CONSEQUENCES OF THE THEOREM OF PYTHAGORAS angle by twice the rectangle contained by one of the sides about the obtuse angle namely that on which the perpendicular falls and the straight line cut off outside by the perpendicular towards the obtuse angle GIVEN AABC LBAC gt right angle CD AB Fig329 TO PROVE BCZ ABZ AC2 2ABAD C D A B Figure 329 PROOF See Exercise 2 PROPOSITION 343II 13 In acuteangled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle namely that on which the perpendicular falls and the straight line cut o within by the perpendicular towards the acute angle GIVEN AABC LBAC lt right angle CD AB Fig330 TO PROVE BCZ ABZ AC2 2ABAD C A D E Figure 330 339 34 CONSEQUENCES OF THE THEOREM OF PYTHAGORAS PROOF See Exercise 3 EXERCISES 34A 1 Explain the relation of Proposition 342 to the Law of Cosines of the trigonometry of the Euclidean plane 2 Prove Proposition 342 3 Explain the relation of Proposition 343 to the Law of Cosines of the trigonometry of the Euclidean plane 4 Prove Proposition 343 Assuming a unit length and a segment of length r units the next proposition deals with the construction of a line segment of length r units The statement that appears here is weaker than that of Euclid s but it is sufficient for this text s purposes and obviates the need for the omitted Proposition 145 Euclid39s version appears in Exercise 2 PROPOSITION 344II 14 To construct a square equal to a given rectangle GIVEN BCDE Fig 331 TO CONSTRUCT Line segment EH such that EH2 2 BCDE Figure 331 CONSTRUCTION If BE 2 ED then BE is the required line segment Otherwise it may be assumed without loss of generality that BE gt ED Extend BE to F so that EF 340 34 CONSEQUENCES OF THE THEOREM OF PYTHAGORAS 2 DE and let G be the midpoint of BF Let H be an intersection of the straight line through E perpendicular to BF with the circle G BG Then EH is the required line segment PROOF By the Theorem of Pythagoras EH2 GHZ GEZ GH GEGHGE BG GEGFGE QED An alternative proof of the equation EH2 2 BE EF appears in Exercise 35C 16 EXERCISES 34B 1 Assume that a line segment of length 1 inch is given Construct a line segment Whose length is a IE inch b E inch c 4 E inch d AVE inch e W inch 2 Prove Euclid39s Proposition 1114 To construct a square equal to a given polygon 5 Proportion and Similarity Euclid39s definition of proportion is too intricate for the context and purpose of this text Instead a shortcut provided by the real number system is used Recall Euclid39s tacit assumption that all geometrical figures have an aspect of size or magnitude Thus lines regions solids and angles have lengths areas volumes and angular measure 341 35 PROPORTION AND SIMILARITY respectively If two figures of the same types have sizes a and 17 relative to some unit it is said that their ratio is the real number 5 The numbers a b c are said to be proportional to the numbers a39 b c39 provided that a E The next five propositions set up some algebraic preliminaries The first of these is of course none other than the Distributive Law and so requires no proof The rest are basic observations regarding proportions PROPOSITION 351V1 If m a b c are any numbers then mambmc mabc are proportional to a39 b c then PROPOSITION 352V12 If a bc abc 1 1 1 a b c a b c 1 1 1 GIVEN a by 2 cy g 2 2 a b c T0 PROVE a b c a b c 1 1 PROOF Let k a by c Then a 2 ka39 1 2 kb39 0 kc abc ka kb kc kabc abc a b c 342 35 PROPORTION AND SIMILARITY QED b i PROPOSITION 353V16 3 ifandonlyif g See Exercise 8 b 39 bb PROPOSITION 354v17 If f then a 7 See Exercise 9 b 39 b b39 PROPOSITION 355V18 If f then M 2 See Exercise 10 EXERCISES 35A In Exercises 177 prove the stated equalities on the basis of the assumption that a b c d are proportional to a39 b39 c39 d39 2 4a 7 3b 20 7d 139 aI 4a39 731239 2039 7d39 2 2 2 2 a 4a 7 3b 20 7512 2 2 2 2 2 2 aI 4a39 731239 2039 7d39 2 2 2 2 ia2bi3c 511 4 2 2 2 2 aI 2b39 73039 5d39 a3 a3 b363 d3 as a3bl3cl3d73 a3b c3d 539 a73b 073d a73c 127311 in SJQ 639 a 3c b 3d 2 2 2 a 3512 c 3b 7 2 2 2 2 a 7 3d 0 7 3b 8 Prove Proposition 353 9 Prove Proposition 354 10 Prove Proposition 356 343 35 PROPORTION AND SIMILARITY PROPOSITION 356VI2 Ifa straight line meets two sides ofa triangle then it is parallel to the third side if and only if it cuts them into proportional segments GIVEN AABC points D and E on AB and CD respectively Fig 332 Q E TO PROVE 1 If DE BC then DB EC AD 2 If m E then DEIIBC Figure 332 Proof of 1 Join CD and BE Then since DE BC A BDE A CDE PN 326 A ADE A ADE A ABE A ACD CN 2 A ABE w A ADE AADE Let EG and DH be altitudes of A ADE Then it follows from the above that ABGE ACDH AD GE AEDH PN 325 Q E AD AE E E AD AE PN 354 344 Proof of 2 35 PROPORTION AND SIMILARITY AD AE EEC Using the same construction as above it is only necessary to reverse the order of the steps of the above argument Exercise 1 QED The proof of the above theorem depends superficially on the additional assumption that the point D lies between A and B Exercises 15 17 rectify this minor flaw EXERCISES 35B 10 ll 12 Complete the proof of Proposition 356 Prove that the straight line that bisects one side of a triangle and is parallel to a second side also bisects the third side Use Proposition 356 twice to prove that the line segment joining the midpoints of two sides of the triangle is parallel to the third side and equals half its length The point K is on the side AB of A ABC points L M are on side AC so that KL H BM and Ail KMHBC Provethat AM AC Point O is not on any of the sides of A ABC or their extensions and K L M are on OA OB OC respectively so that KL H AB and LMHBC Prove that KMH AC Prove that if D is any point on the side BC of A ABC then AD bisects the interior angle at A 2 if and only if AC DC Prove that if E is any point on the extension of side BC of AABC then AE bisects the exterior AB BE angleat A 1fandonly1f E E Prove that if the straight lines m1 m2 m are all parallel to one side of a triangle and they cut off equal segments on a second side then they also cut off equal segments on the third side Divide a given line segment into three equal parts Let n be a given positive integer Divide a given line segment into 11 equal parts Let m and n be given positive integers Divide a given line segment into two parts whose ratio is mn Let a b c be three given line segments Divide a into two parts whose ratio equals bc 345 13 14 15 16 17 18 19 20 21 22 23 24C 35 PROPORTION AND SIMILARITY Let a b c be three given line segments Construct a line segment x such that X a Let a b be two given line segments Constructa line segment x such that g b Show that the proof of Proposition 356 still holds with minor modifications when A lies in between B and D Show that the proof of Proposition 356 still holds with minor modifications when B lies in between A and D Does the proof of Proposition 356 require any other corrections Comment on Proposition 356 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi Use the spherical trigonometry formulas to experiment with comparing the length of the line joining the midpoints of two sides of a spherical triangle with that of the third side see Exercise 2 above Form a conjecture regarding the relative sizes of these two geodesic segments Use the hyperbolic trigonometry formulas to experiment with comparing the length of the line joining the midpoints of two sides of a hyperbolic triangle with that of the third side see Exercise 2 above Form a conjecture regarding the relative sizes of these two geodesic segments Comment on Exercise 19 in the context of taxicab geometry Comment on Exercise 19 in the context of maxi geometry Construct AABC given the following data a 1 mb me b ma mb me Use a computer application to verify part 1 of Proposition 356 Similar polygons are those whose corresponding angles are equal and whose corresponding sides are proportional Congruent triangles are similar Exercise 2 and the relation of similarity is transitive Exercise 3 If A ABC and A DEF are similar this is denoted by A ABC A DEF where it is implicit that A B C correspond to D E F respectively as was the case for congruent triangles The next proposition is known as the AAA similarity theorem PROPOSITION 357VI4 Equiangular triangles are similar GIVEN AABC A DEF L ABC 2 A DEF A ACB A DFE A BAC A EDF Fig 333 346 BC 35 PROPORTION AND SIMILARITY TOPROVE DE 2 EF 2 BF D 5 G A E F Figure 333 PROOF If any side of A ABC equals the corresponding side of A DEF then the two triangles are congruent ASA Hence it may be assumed without loss of generality that AB lt DE Let G be a point in the interior of DE such that DG 2 AB Let H be the intersection of the straight line parallel to EF through G with DF Then DG E DG E DH DH PN 356 PN 355 However AABC s A DEF by ASA because 42 AB 41 AC AB DE A similar argument can be used to prove that 42 DG 4 4 DH DG DE Both equal 4 3 Construction Given DF E E EFZDF QED 347 35 PROPORTION AND SIMILARITY The following similarity theorems are known as the SSS and SAS similarity theorems not to be confused with the SSS and SSA congruence theorems and their proofs are relegated to Exercises 5 and 6 respectively PROPOSITION 358VI5 If two triangles have their sides proportional then the triangles are similar PROPOSITION 359VI6 If two triangles have one angle equal to one angle and the sides about those angles are proportional then the triangles are similar In view of the fact that the sum of the angles of every Euclidean triangle is 1800 it follows that the conclusion of Proposition 357 holds even when only two of the angles of one triangle are known to be equal to the corresponding angles of the other triangle EXERCISES 35C W W QMP 10 Complete the proof of Proposition 357 Prove that congruent triangles are similar Prove that if AABC is similar to AA39B39C39 and AA39B39C39 is similar to AAquotBquotCquot then AABC is similar to A AquotBquotCquot Prove Proposition 358 Prove Proposition 359 Prove that in similar triangles corresponding altitudes are proportional to corresponding sides Prove that in similar triangles corresponding medians are proportional to corresponding sides Prove that in similar triangles the corresponding angle bisectors are proportional to the corresponding sides Prove that the areas of similar triangles are proportional to the squares of their corresponding sides Prove that the areas of similar polygons are proportional to the squares of their corresponding sides V119 348 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 35 PROPORTION AND SIMILARITY Three parallel straight lines cut the straight lines 711 and n in the points A B C and K L M AB KL respectrvely Prove that if m In N ABCD the straight line BKLM cuts the diagonal AC in the point K and the possibly BK KM extended sides AD and CD in the points L and M respectively Prove that E y In N ABCD the straight line BKN cuts the possibly extended sides CD and AD in the points AD CK K and N respectively Prove that m E Prove that the intersection point of two of a triangle39s altitudes divides them so that the product of each altitude39s segments equals the product of the other39s segments Prove Euclid39s Proposition V18 If in a right tiiangle a perpendicular be drawn from the right angle to the opposite side then the triangles so formed are similar to each other and to the given triangle Prove Euclid39s Proposition V113 The square of the altitude to the hypotenuse of a right triangle equals the product of the segments it determines on the hypotenuse Prove Euclid39s Proposition V131 If similar polygons are constructed on the sides of a right triangle then the polygon on the hypotenuse equals the sum of the polygons on the other two sides Describe the locus of all the points whose distances from the sides of a given angle have a ratio equal to that of two given line segments Given an angle and a point A inside it construct through A a straight line whose portion between the sides of the angle is divided by A into segments whose ratio equals that of two given line segments Given an angle and a point A inside it find a point P on one side of the angle whose distance from A equals its distance from the other side of the angle How many solutions are there Prove that the perimeters of similar triangles are proportional to their corresponding sides Prove that the perimeters of similar polygons are proportional to their corresponding sides A straight line through the intersection of the diagonals of a trapezoid is parallel to its parallel sides Prove that the segment between the noniparallel sides is bisected by the intersection of the diagonals Prove that in a trapezoid the line joining the midpoint of one of the parallel sides to the intersection of the diagonals bisects both the parallel sides Prove that in a trapezoid which is not a parallelogram the straight line joining the intersection of the diagonals to the intersection of the noniparallel sides bisects both the parallel sides Comment on Proposition 357 in the context of the following geometries a spherical b hyperbolic c taxicab d maxi 349 CHAPTER REVIEW CHAPTER REVIEW EXERCISES 10 11 12 13 14 15 Let P be a point in the interior of the equilateral A ABC Prove that the sum of the perpendicular segments from P to the sides of the triangle is constant The diagonals of a trapezoid cut each other into segments that are proportional to the parallel sides of the trapezoid Suppose the noniparallel sides of a trapezoid are equal Prove the following i The two angles adjacent to the same base are equal ii The diagonals are equal iii The diagonals intersect in a point that lies on the straight line joining the midpoints of the unequal sides iv The diagonals divide each other into respectively equal segments v The midpoints of the four sides form a rhombus Find a simple expression for the angle between two of a triangle39s angle bisectors A straight line through the verteX of a triangle divides it into two triangles that are similar to each other and to the original triangle Prove that the given triangle is a right triangle In NABCD a straight line parallel to AB intersects AD AC and BC in the points P Q R respectively Prove that AAPR AAQD In AABC ma BC2 Prove that A CAB is a right angle In AABC A BAC 2 4 ABC Prove that a2 22 c In NABCD M and N are the midpoints of the opposite sides AB and CD Prove that the straight lines DM and BN divide the diagonal AC into three equal segments Suppose that A ACB of A ABC is obtuse and the perpendicular bisectors to AC and BC intersect AB in the points D and E respectively Prove that LDCE 24 ACB 7 900 From a point on the base of an isosceles triangle straight lines parallel to the triangle39s other sides are drawn Prove that the perimeter of the parallelogram thus formed is independent of the position of the point on the base Prove that the bisectors of the two angles formed by the opposite pairs of sides of a conveX quadiilateral intersect in angle that equals half the sum of two opposite angles of the quadiilateral Prove that the median and the altitude to the hypotenuse of a right triangle form an angle that equals the difference of the triangle39s other two angles In AABC AB 2 AC E and D are on the sides AB and AC respectively and LABD 200 LCBD 60 LBCE 50 and LACE 30 Find LEDB Are the following statements true or false Justify your answers a Playfair s postulate is valid in neutral geometry 350 b c d e f g h i l k 1 m n 0 P CHAPTER REVIEW Playfair s postulate is valid in Euclidean geometry Playfair s postulate is valid in spherical geometry Playfair s postulate is valid in hyperbolic geometry Playfair s postulate is valid in taxicab geometry If in a quadrilateral one pair of opposite sides are equal as are one pair of opposite angles then the quadrilateral is a parallelogram There is a Euclidian right triangle with sides 287 816 865 There is a neutral right triangle with sides 287 816 865 There is a spherical right triangle with sides 287 816 865 There is a hyperbolic right triangle with sides 287 816 865 There is a taxicab right triangle with sides 287 816 865 If the corresponding sides of two triangles are proportional then so are their corresponding angles Equiangular triangles are similar The corresponding sides of equiangular quadrilaterals are proportional If the corresponding angles of two quadrilaterals on the surface of a sphere are equal then so are their areas Equiangular quadrilaterals are similar 351
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