A fascinating application of Steiner’s Theorem for Trapezium: Geometric constructions using straightedge alone
Moshe Stupel Shaanan Academic Religious Teachers College, Haifa, Israel
ased on Steiner’s fascinating theorem for trapezium, a seven geometrical Bconstructions using straightedge alone are described. These constructions provide an excellent base for teaching theorems and the properties of geometrical shapes, as well as challenging thought and inspiring deeper insight into the world of geometry. In particular, this article also mentions the orthic triangle and proves its special property, and shows some other interesting constructions, such as, for example, how to construct a circle’s diameter using straightedge alone and having only a segment with its midpoint. In addition, it is enhanced by aspects of the historical background of geometric constructions, including reference to “impossible constructions.” Application of the material presented in college or high school can enhance students’ appreciation of the elegance, beauty, and fascination of mathematics. Through such “adventures,” students will be encouraged to further pursue geometric problems and explore various methods of problem solving, especially those concerned with geometric constructions.
Introduction
The motivation for this article stems from the presentation of a geometrical problem (from a high school textbook) to pre-service mathematics teachers during a course in plane geometry given in an Israeli teacher’s college for prospective high school teachers. The purpose of the course was to deepen their knowledge in this topic, and included the construction of geometrical forms incorporated with the history of geometric constructions. The problem presented was the following: The sides of trapezium ABCD meet at point F. A line segment, FN, passes through the intersection of the diagonals, E (see Figure 1). Prove that AM = MB and that DN = NC. Note: A trapezium is a quadrilateral with one pair of opposite sides parallel. Australian Senior Mathematics Journal vol. 27 no. 2
6 A fascinating application of Steiner’s Theorem for Trapezium Australian Senior Mathematics Journal vol. 27 no. 2 7 Figure 1. A trapezium ABCD with FN passes through the intersection point of the diagonals, E. the intersection point of the diagonals, FN passes through 1. A trapezium ABCD with Figure The aim of this article is to point out the value of developing pre-service The students noted that Steiner’s proof was comparable to the solution of The students noted that Steiner’s The problem is solved by building an auxiliaryThe problem is (which generally is figure sides, immediately reminded the authors of Steiner’s theorem, which in fact in which theorem, Steiner’s of authors the reminded immediately sides, above. states the results of the problem a straightedge only. a straightedge only. (not to be confused with confused be (not to theorem as Thales’ known also , theorem the intercept and elementaryin theorem important an is It name). that with theorem another geometry if two about the ratios of various line segments that are created fascination with mathematics. intersecting lines are intercepted by a pair of parallels. It is equivalent to intersecting lines are intercepted bases, the meeting point of the trapezium’s passes through the midpoints of of the development of geometric constructions to increase their interest and of the development of geometric constructions to increase their constructions in general along with their ability to construct geometric figures constructions in general along with their ability to construct geometric continuation of the sides—are on the same line. which continue researching the use of the Steiner theorem for the trapezium, base, the point where the diagonals cross, and the point of meeting of the base, the point where the diagonals cross, and the point of meeting using only a straightedge, along with giving them the historical perspective using only a straightedge, along with giving them the historical teachers’, in-service teachers’, and students’ understanding of geometric ultimately led to an interest in general geometric constructions according to ultimately led to an interest in general geometric constructions according using the rules of ancient Greek mathematics, and building geometric figures their problem (the proof of which is given below) and thus were stimulated to were and thus below) is given which of (the proof problem their that the segment FN triangles. However, the theorem about ratios in similar continuations of the of the trapezium’s the diagonals and the meeting point the key to the solution)—passing a line parallel to the base of the trapezium the key to the solution)—passing (Jakob Steiner, 1796–1863) (Jakob Steiner, For every trapezium ABCD, the following four points—the midpoints of each ABCD through the meeting point of the diagonals, using similar triangles, ABCD through the meeting point The Steiner Theorem for Trapezium The Steiner Theorem for Trapezium 8 Australian Senior Mathematics Journal vol. 27 no. 2 Stpel & Ben-Chaim 7 5 n 5537).Gaussprovedthatann 17, 257and65 1939) provedtheimpossibilityofthirdproblembyshowingthatsinceπis 1981): 1848) asindicatedbySuzuki(2008).In1837,Wantzel provedtheimpossibility 7-gon isnotsuchanumber, aswell9-gonor 11-gon. The rulesofancientGreekmathematicsallowgeometricconstructionsusing This popularitymaypossiblybeattributedtotheancientGreeks’introduction 2, and,possibly, multipliedbydistinctFermatprimes. Obviously, theregular 2001). HisproofinvolvedtheFermatprimes(the onlyonesknownare3,5, as a root (Berggren, Borwein & Borwein, 1997),andconsequently, &Borwein, π asaroot(Berggren,Borwein squaring heritage andhavebeenapopularpartofmathematicsthroughouthistory. without markings,andwithonlyoneedge.Acompasscanbeusedtoinscribe the circleisimpossible. transcendental, noequationofanydegreewithrational coefficientscanhave they werefinallyproventobeimpossibleinthenineteenthcentury. Theproofs they were not able to actuallyprove their impossibility, and these problems constructed by compass and straightedge if and only if n earlier, in1790,byCarlFriedrichGauss(Childs,2009;Krizek,Luca&Somer, circles, butcannotbeusedtotransfermeasurements. only acompassandstraightedge.Astraightedgeisdefinedasinfiniteinlength, of thefollowingfourfamousgeometricconstructionswhichhavebeenproven investigations of these impossible geometric constructions, mathematicians of degree2 angleusinghistheoremthat of duplicatingthecubeortrisectinganarbitrary of impossibility of two of the constructions came from Pierre Wantzel (1814– impossible toconstructmanyhundredsofyearslater(see,forexample,Heath, important. First,geometricconstructionsareapartofthemathematical Importance ofgeometricconstructions Importance if drew the attention of many famous mathematicians over the centuries, until at thatsametime.Then,in1882,KarlFerdinandVon Lindemann(1852– • • • • r Trisecting theangle:Givenanyangle,findageometricconstructionforan Inscribing aregular heptagon inacircle: Constructaregularheptagon(seven- Although theGreeksdidbelievethattheseconstructionswereimpossible, Doubling thecube:Givenanycube,findageometricconstructionforcube Squaring thecircle: Givenanycircle,findageometricconstructionfor isaconstructiblenumber, itmustbetherootofanirreduciblepolynomial The impossibilityofconstructingaregularheptagon wasactuallysolved There are several reasons why geometric constructions are considered However, whatisevenmoreimportanttounderstand isthat,throughtheir with twicethevolumeofgivencube. angle thatisone-thirdthemeasureofgivenangle. sided polygon)withcompassandstraightedgeonly. square ofequalareatothecircle. n . In essence, this is the basis of the Galois Theory thatwasderived . Inessence, this isthebasisof Galois Theory -sided regularpolygoncanbe is equal to apower of A fascinating application of Steiner’s Theorem for Trapezium Australian Senior Mathematics Journal vol. 27 no. 2 9 The authors of this manuscript believe that geometric constructions, The authors of this manuscript The reason for exploring the history argued of geometric constructions is students in the upper grades of secondary schools. In fact, trying to learn and geometrydeveloping ideas of ‘proofs’. In addition, can be useful in and originality, while presenting opportunities for integrating solutions and while presenting opportunities for integrating and originality, arrived trying attempt to invent inspired them to above problems to solve the tangible” (Robertson, experience, something a Piagetian concrete-operational geometry is like trying without geometric construction to learn chemistry or geometry. Geometric constructions expand the student’s comprehension of constructions expand the student’s Geometric geometry. curriculum Incorporating geometric construction into the inspirational part of mathematics tending to engender mathematical thinking, inspirational part of mathematics tending to engender mathematical Reporting Authority and Assessment Curriculum, Australian the inspecting developing unique strategies. Indeed, when dealing with a construction developing unique strategies. Indeed, emphasise the beauty of mathematics are problem, interesting solutions that development of geometry. For example, the dead end to which the Greeks which the Greeks dead end to For example, the of geometry. development is definitely important to learn” (p. 1). Swetz (1995) also believes that historyis definitely important to learn” deals with science, social studies, or mathematics, its historydeals with science, social studies, should be promote true problem solving through the use of reasoning” (p. 36). solving through the use of promote true problem constructions, their incorporation into geometryincorporation their constructions, is veryinstruction important, especially for pre- and in-service mathematics teachers and for high-level often revealed. Complicated construction problems constitute a powerfuloften revealed. Complicated construction development, and allows the student challenge which contributes to thought constructions. Second, “geometric constructions can reinforce proof and can reinforce proof constructions Second, “geometric constructions. mathematics. can supply the why, where, and how for many concepts that are studied. where, and how for many concepts that can supply the why, explored. Specifically, to fully understand geometric constructions, the history explored. Specifically, more technologically advanced tools to enable them to perform advanced tools to enable them more technologically the required by Lamphier (2004) who said that “to fully understand a topic, whether it a topic, whether fully understand “to that said who Lamphier (2004) by the school mathematics curriculum all over the world. Geometry is an geometric topic of geometric constructions is included. Given the value of to implement the important properties of known geometric shapes in the to implement the important properties through the various methods of their solution, encourage thought, creativity through the various methods of that “unless constructions simply ask students to mimic a given example, they simply ask students to mimic that “unless constructions lends visual clarity to many geometric relationships” (Sanders, 1998, p. 554). p. 1998, (Sanders, relationships” geometric many to clarity visual lends were able to achieve many remarkable feats that contributed much to the much to feats that contributed remarkable to achieve many were able Euclidean or projective plane, and thus diversify the process of studying Euclidean or projective plane, The topic of geometryserving is always topics across as one of the major Third, geometric constructions “give the secondaryThird, geometric school student, starved for 1986, p. 380). Another reason is presented by Pandisico (2002) who claimed reason is presented by Pandisico 1986, p. 380). Another (ACARA) documents, in each geometry program for high school grades, the 10 Australian Senior Mathematics Journal vol. 27 no. 2 Stpel & Ben-Chaim “digital technologiescanmakepreviouslyinaccessiblemathematicsaccessible, Three types of geometricconstructions Three Shape of the Australian Curriculum: Mathematics(May The ACARAdocument,ShapeoftheAustralianCurriculum: There arethreetypesofgeometricconstructions, based onthetoolsthatare It statesthat“digitaltechnologiesallownewapproachestoexplainingand 2009) emphasises the roleofdigitaltechnologies in teachingmathematics. when onereferstobuildingthebasicstructures. without using any measurements, and is the construction method implied with theirhistoricalcontextsenlivensthetopic.Thebasicknowledgeandskills relationships (Lamphier, 2004). theorems. Geometric construction can also help teachers transform the geometricconstructionsencouragestudentstodiscover needed toperform to create very complex and yet very precise geometric constructions and complex and yet very to create very unit values),compassonly, andstraightedgeonly. Thefirsttypeistheoriginal, used toconstructthem:compass-and-straightedge (with orwithoutrelatingto biology withoutlaboratories.Presentinggeometricconstructionstogether more students,includingtheuseofrealisticdataandexamples”(6.5.2).Infact, constructions. However, onemustbeawarethatwhileDGShelpstovisualise currently, theintroductionofDGSandrapidprogresstechnology classic constructionmethodusingonlyacompass andstraightedge, of fivebasicconstructionmethodsusingpoints, lines,andcirclesthathave is dynamicgeometricsoftware(DGS)suchasTheGeometer’s SketchpadorGeogebra. diagrams. Obviously, static constructions lack the strongimpactof dynamic presenting mathematics,aswellassistinginconnectingrepresentations definition, postulatesandtheorems),studentswhouseDGScannotbe geometry textbooks into an active and exploratory investigationofgeometric textbooksintoanactiveandexploratory geometry geometric constructionswithintheframeworkoftheirmathematicscurriculum and enhance the potential forteachers to make mathematics interesting to and thusdeepeningunderstanding”(6.5.1).Then,continuestoindicatethat and exploregeometricrelationshipsinterpretconcepts allowed toignorelearningandpresentingformalproofsfortheproblems. a relationship,itdoesnotprovideformalproof(usingappropriategeometry and its availability in education, produces an environment that enables both already beenconstructed. Thesemethodsare: static andconfusingarrayofdefinitionstheoremstypicallyfoundin students andteacherstoexploregeometricrelationshipsdynamically • • • A very helpfultoolforencouragingandmotivatingteacherstoincorporate A very All compass-and-straightedge constructions consist of repeated application creating apointwhichis the intersectionoftwoexisting,non-parallellines; creating acirclethrough onepointwithcentreanotherpoint; creating alinethroughtwo existingpoints; A fascinating application of Steiner’s Theorem for Trapezium Australian Senior Mathematics Journal vol. 27 no. 2 11 and the . x d n a y x , y ⋅ x , y − x , y + x , and a unit length of 1. Through basic geometry, and a unit length and algebra, Solving construction tasks using only a straightedge requires a wider and Solving construction tasks using only a straightedge requires a wider intersect); and intersect); creating one or two points at the intersection of two circles (if they intersect). of two circles at the intersection one or two points creating creating one or two points at the intersection of a line and a circle (if they a circle (if they of a line and at the intersection one or two points creating Geometric constructions using only a straightedge is a topic in plane Geometric constructions using only a straightedge is a topic The second type of geometric construction uses compass alone. According uses compass alone. The second type of geometric construction with straightedge alone. In this case, the The third type of construction is Another approach to basic constructions is by relating to the ‘numbers’ to basic constructions is by Another approach • • solutions are revealed that emphasise the beauty of mathematics. solutions are revealed that emphasise the beauty of mathematics. straightedge alone, and most of the constructions presented are conducted straightedge alone, and most of the constructions presented are Geometric constructions with straightedge alone arithmetic constructions for two given segments, one of length x for two given segments, arithmetic constructions alone provided one is given a single circle and its centre. Actually, this means circle and its centre. Actually, alone provided one is given a single given a circle and a line passing through its centre. In other words, using a In other centre. its through a line passing circle and given a geometry in the program of studies (which is that is not directly expressed problem, especially with only a straightedge (or only a compass), interesting problem, especially with only a straightedge (or only a compass), possible: proved this theorem in the second volume of his writings: “Die geometrischen proved this theorem in the second circle and a diameter, but without knowing the precise location of its centre. but without knowing circle and a diameter, other length y arithmetic constructions are could be constructed. Five other related lengths compass alone. In this case, a straight line is defined by any pair of points. compass alone. In this case, a straight more profound knowledge of plane geometry than that required to solve one action with the compass to provide the circle and its centre. Steiner centre. its and circle the provide to compass the with action one by compass and straightedge together can be constructed by straightedge by compass and straightedge together that the ancient Greeks were able to construct with compass and straightedge Greeks were able to construct with that the ancient unfortunate, given that it encourages thought, creativity and originality in unfortunate, given that it encourages thought, creativity and originality solutions the methods of solution, and presents opportunities for integrating to the Mohr–Mascheroni theorem (Eves, 1968), any geometric construction to the Mohr–Mascheroni theorem that can be performed a with a compass and straightedge can be done with line from a point either distant to or on a line using only a straightedge, and line from a point either distant to or on a line using only a straightedge, while actually doing arithmetic geometrically by lengths of segments. Knowing Knowing segments. of lengths by geometrically arithmetic doing actually while while developing unique strategies). Often, when dealing with a construction while developing unique strategies). Often, when dealing with a construction without being given the centre of the circle. In particular, we shall focus on without being given the centre of the circle. In particular, how to construct a parallel line to a given line, they were capable of making a parallel line to a given line, they how to construct how to construct a perpendicular (and occasionally a parallel) to a straight how to construct a perpendicular (and occasionally a parallel) to konstruktionen ausgeführt mittels der geraden linie, und eines festen kreises”. konstruktionen ausgeführt mittels Poncelet–Steiner theorem (Eves, 1995) states that whatever can be constructed states that (Eves, 1995) theorem Poncelet–Steiner
This article focuses on the third type of geometric construction: that with This article focuses on the third type of geometric construction: 12 Australian Senior Mathematics Journal vol. 27 no. 2 Stpel & Ben-Chaim (including theonespresentedinAustralian schools), whenthisproblem P A circlewithdiameterABisdrawnonaplane,aswellsomerandom Presentation ofthetask In addition,thesolution requiressomeknowledgeofthefollowingspecial learned in plane geometry aspartofthehigh-school programofstudies learned inplanegeometry requires solutionsforallsevencases. them duringpresentation oftheproblem. the circleandspecialsegmentsintriangle(altitudes, anglebisectors,etc.). or projectiveplane,andthustodiversifytheprocessofstudyinggeometry. complexity oftheproblem,anditssolution,dependsonlocationpoint properties oftrapeziums (ofwhichthestudentsareusuallynotaware),such is presentedtostudents,itshouldbe only aftertheyhavestudied problems usingacompassandstraightedge.Suchconstructionsallowoneto perpendicular todiameterABisbeconstructed,usingastraightedgeonly. point denotedbytheletterP as theSteinertheoremfor atrapezium,andwhichshouldbeintroduced to apply the important properties of known geometric shapes in the Euclidean • • • • • • • relativetolineABandthegivencircle,fullsolutionoftask Case 4:PointP Case 3:PointP Case 7:PointP Case 6:PointP Case 1:PointP Case 5:PointP Case 2:PointP The sevenpossiblelocationsareasfollows: There aresevenpossiblelocationsofpointP Because thesolutionoftaskrequiressomeknowledge ofthematerial with theendsofdiameter line ABliesonthediameter the diameter outside thecircle. continuation of . =P =AP =P =P =P =P =P AB liesoutsidethecircle. 1 Figure 2.Thesevenpossiblelocations forpointP. 5 3 4 islocatedoutsidethecircle,sothatitsprojectionon 2 isaninteriorpointonthediameter. isaninteriorpointofthecircle,whichdoesnotlieon islocatedonthecirculararc,anditdoesnotcoincide islocatedoutsidethecircle,anditsprojectionon 7 islocatedonthecontinuationofdiameter AB, 6 islocatedattheendA . AstraightlinepassingthroughpointP AB. A andB . ofthediameter. (seeFigure2),andthe and A fascinating application of Steiner’s Theorem for Trapezium Australian Senior Mathematics Journal vol. 27 no. 2 13
KL and F , M , and the segment and the is a median to the side the to median a is EF at the point N DC at parallel to base AB. From the theorem , that is, that EL, = KE , it will be proven that the points E DN = NC, it will be proven . denote the middles of the bases of trapezium ABCD. Figure 3. Line EF intersects the bases of the trapezium at the points M and N. 3. Figure (Figure 3). and N KL bisects any segment parallel to KL, whose ends are located on the Steiner’s proof of the trapezium theorem: To prove the theorem, it will be prove the theorem, of the trapezium theorem: To proof Steiner’s E Draw segment KL through point We use Steiner’s theorem to solve our task. At the end of the article is use Steiner’s We This theorem is applied for the two triangles ∆FKL and ∆FDC, and then The continuations of the sides of two trapeziums with same height, a height, with same trapeziums two of sides the of continuations The Another theorem is also used: “In any triangle ∆FKL, the median to the lie on the same straight line, as shown on Figure 1. straight line, as shown on Figure lie on the same side sides of the angle ∠KFL”. (This theorem can also be proven using Thales’ another example for a geometrical problem that can be solved simply using another example for a geometrical problem that can be solved simply and N Steiner’s theorem for the trapezium, as well as an application of the presented Steiner’s parallel to the base whose ends are on the sides of the trapezium and which parallel to the base whose ends are be proven using Thales’ (can of intersection” point through the passes in triangle ∆FKL. is stated in the introduction, above. While many other mathematicians have have many other mathematicians above. While in the introduction, is stated proof.) Steiner’s only to reference is there article in this theorem, this proven intersects the bases of the trapezium at the points of the trapezium at the pointsM line EF intersects the bases proven that the intersects the segment that FE intersects obtaining common large base, and a small base with a similar length, intersect at points common large base, and a small base with a similar length, intersect constructions. Property theorem The Steiner the trapezium. (Note: theorem for A: The Steiner Property B: Is a result of the Steiner theorem for the trapezium. theorem), it is concluded that concluded is it theorem), theorem). which are located on one line parallel to the bases of the trapeziums. Let M MB, Given that: AM = Special properties of the trapezium of the Special properties N AB at the point M “the point of intersection of the diagonals of a trapezium bisects the segment “the point of intersection of the 14 Australian Senior Mathematics Journal vol. 27 no. 2 Stpel & Ben-Chaim Constructing anapplicationoftheSteinertheorem forthetrapezium(will be Constructing The application:To constructastraightlineparalleltogivensegmentthat On thecontinuationofline,choosesomepointX whose continuationintersectsBX used below) large baseandequalsmallbases.PointsG triangles: ∆GDN~GMB,andtheinverseofThales’theorem. that isparalleltoABandpassesthroughpointP : Anindirectproofisgiven.Supposethatline PX Proof oftheconstruction : DrawastraightlinethroughpointsA Description oftheconstruction passes throughagivenpoint,whenthemidpointofsegmentisgiven;i.e., points parallel toAB(Figure6). is notparalleltoAB.DrawthroughthepointP parallel tothebasesoftrapeziums. given apointP and Figure 4.Two trapeziums:MBNDandAMND,whichhaveacommonlargebaseequalsmallbases. The proofemploysthesimilarityoftriangles:∆FNC~FMBand Figure 4showstwotrapeziums:MBNDandAMND,whichhaveacommon CX B 1 intersect at point andC (Figure5).ConnectpointB and asegmentABwithmidpointC Figure aparallellinethrough 5.Descriptionofconstructing a pointPtogivensegmentABanditsmidpointC. X 2 . Draw a straight line connecting points 1 atpointX andF 3 withpointP . ThelinePX thestraightlinePQ,whichis areononestraightline , constructastraightline . 1 . ConnectpointX . ThesegmentsBP 3 isthesoughtline. A and andP 1 with X 2 3
. , A fascinating application of Steiner’s Theorem for Trapezium Australian Senior Mathematics Journal vol. 27 no. 2 15 are also 1 CC , respectively. respectively. , , 1 BB , and R 1 2 and AQ are different , as shown in Figure 7. 3 1 C 1 B 1 ∆A intersects base AB, which means R 1 , respectively, which means that the , respectively, || AB. 3 are also different lines, and they intersect are also different lines, and they and D at different points X at different BP R 1 do not coincide, lines AX do not coincide, and X 2 X 1 and Q 3 through a point P to a given segment AB and its midpoint C. a point P to a given segment through Figure 6. Description of indirect proof to a construction line of a parallel proof 6. Description of indirect Figure AB at different points C . On the other hand, from Steiner’s . On the other hand, from Steiner’s is not the middle of segment AB D . Hence, we have a contradiction! Therefore, must be the midpoint of AB. Hence, we have a contradiction! D angle in a semicircle is a right angle). Since points X given a segment and a parallel line to the segment, bisect the segment. given a segment and a parallel line given a segment and its middle point, construct a parallel to the segment given a segment and its middle point, concurrent at a point), called the orthocentre of the triangle. midpoint of the arc that corresponds to the chord), is perpendicular to midpoint of the arc that corresponds to the chord), is perpendicular equal). the chord. through any point; and the bisectors of the angles in the triangle In a triangle ∆ABC, the three heights (altitudes) AA Inscribed angles that rest on the same arc (or on equal arcs) are equal Inscribed angles that rest on the same arc (or on equal arcs) are The three heights (altitudes) of a triangle intersect at one point (are The three heights (altitudes) of a triangle intersect at one point The description of this construction is the general process for constructing constructing process for the general is construction this of The description A diameter that is perpendicular to a chord bisects it. A diameter that passes though the midpoint of a chord (or through the of a midpoint the though passes A diameter that An inscribed angle that rests on the diameter is a right angle (ACARA: An An inscribed angle that rests on the diameter is a right angle (ACARA: (ACARA: Two angles at the circumference subtended by the same arc are angles at the circumference subtended by the (ACARA: Two • • • • • • • • segment a line parallel to a given segment (with its midpoint). It is important to note to important is It midpoint). its (with segment given a to parallel line a point constructions with a straightedge alone are equivalent: constructions with a straightedge the assumption is not correct, and PX the assumption is not correct, and that X theorem, in trapezium ABPQ, the line that according to the Steiner theorem for the trapezium, the following two that according to the Steiner theorem lines, and they intersect segment they intersect and lines, task: The properties of the heights in a triangle: The properties of a diameter and a chord: The properties of angles in a circle: Therefore, the lines X A list of other geometrical properties which are relevant for the solution of the of solution the for relevant are which properties geometrical other of list A 16 Australian Senior Mathematics Journal vol. 27 no. 2 Stpel & Ben-Chaim = γ. = & Toeplitz, 1957).To provethepropertyoforthictriangle,refertoFigure 9. Fagnano’s problem,posedin1775,offindingfortheminimumperimeter hence point therefore theheightsof theoriginaltriangle∆ABCarebisectorsof triangle triangle (seeFigure8).Also,theorthicprovidessolutionto triangle inscribed in a given acute-angle triangle (Holand, 2007; Rademacher orthic triangle∆A obtain in thecircletoobtain,accordingsumsofoppositeangles,∠ABC inscribed circle)oftheorthictriangleisorthocentreoriginal similar toeachotherandtheoriginaltriangle, ∆ABC.Ateachvertexof ∠AB ∆ABC. The resultshowsthatthethreetriangles,∆A Triangle It isinterestingtoindicatethattheincentre(thatcentrefor Inscribe 1 ∠BAC = C ∆A 1 =β.Similarly, circumscribe quadrilateral AB ΔBCB 1 ∆A B Figure 8.Thecentre triangle∆A oftheinscribedcircle oftheorthic C 1 1 C 1 isonthecircle. Therefore, quadrilateralBCB ∠CA B 1 aretwoequalanglesof90°–α,βand γ, and Figure 7.Thethree heightsAA 1 1 are alsothebisectorsofangles inthetriangle∆A 1 C with a circle. Obviously, B 1 1 orthic triangleoraltitudeof iscalledtheorthic 1 C B 1 1 . = is the orthocentre oftheoriginaltriangle. is theorthocentre α. Finally, isolate 1 , BB BC is the diameter of this circle and A 1 , CC 1 C 1 1 AC to obtain 1 oftriangleABC BC 1 , ∆AB 1 B 1 A 1 1 C C 1 1 B . 1 and∆A 1 withacircle,to ∠ACB = B 1 1 C C 1 1 , isinscribed 1 B ∠AC 1 C, are 1 B 1
A fascinating application of Steiner’s Theorem for Trapezium Australian Senior Mathematics Journal vol. 27 no. 2 17
,
1 2 B 1 intersect at 1 is connected with point X 1 is the point of . 3 in direction C 1 1 C 1 and BX B 2 1 , since the angle of impact of angle the since , are heights in the triangle 1 1 A 1 B . Connect point A 1 2 C and BX 2 and X 1 . The two segments AX 1 using a straightedge only (see Figure 2), it is Figure only (see straightedge a using . 1 P , whose continuation intersects the diameter at point 3 X 1 with point X (the ends of the diameter). These lines intersect the Figure 9. Proof of the property of the orthic 9. Proof triangle ∆A Figure and B is located at point P . The line P 3 (inscribed angles resting on the diameter), and X angles resting (inscribed X B 1 Now that we have presented all the geometric statements required for Now that we have presented all It is interesting to note that in high school geometryIt is interesting to note that in high learn studies, students This property of the orthic triangle (see Figure 7) also has a physical the orthic triangle (see Figure This property of , is the requested perpendicular. , is the requested perpendicular. straightforward manner. straightforward manner. significance: any ray of light that penetrates through point C significance: any ray of light that and connect point B about the special points related to the points of intersection of the three mid- about the special points related to an ‘optical’ or ‘billiard’ path reflecting off the sides of ΔABC. an ‘optical’ or ‘billiard’ path reflecting point possible to proceed. passing through given point given through passing perpendiculars, the three angle bisectors, the three medians, and the three the three angle bisectors, the three medians, perpendiculars, inside), is equal to the angle of return. The sides of the orthic triangle form triangle orthic the of sides The return. of angle the to equal is inside), circumference of the circle at points X continues to travel indefinitely along route route along indefinitely travel to continues of the inscribed circle in the orthic triangle. of the inscribed circle in the orthic of the circle circumscribed around the triangle, the second is the centre of of the circle circumscribed around of the light on each side (assuming it is coated with a mirror surfaceof the light on each side (assuming the from Proof of the constructionProof AX : Segments Description of the construction: As shown in Figure 10, point P to points A the completion of the task, that is, to find the perpendicular to diameter AB the completion of the task, that though, one can see that the point of intersection of the heights is the centre though, one can see that the point the inscribed circle, and the third is the centre of gravity. However, they are However, is the centre of gravity. the inscribed circle, and the third not given any special property for the fourth one, the orthocentre. From here, From orthocentre. the one, fourth the for property special any given not heights in a triangle. They learn that the first intersection point is the centre heights in a triangle. They learn Case 1: P Solution of the task There are seven different solutions, depending on the location of point P There are seven different solutions, depending on the location ∆AP M (see Figure 2), plus another option using the Steiner theorem in a more (see Figure 2), plus another option using the Steiner theorem 18 Australian Senior Mathematics Journal vol. 27 no. 2 Stpel & Ben-Chaim Therefore, thepointofintersectionheightsisX Case 3:P Case 2:P first case,andasshowninFigure11. first case,andasshowninFigure12. height tothesideAP the triangle∆AP : Thestagesoftheconstructionareasin Description oftheconstruction : Thestagesoftheconstructionareasin Description oftheconstruction : Inthiscase,triangle∆AP Proof oftheconstruction intersection oftheheightsintriangle,thereforeP in thetriangle. Figure ofaperpendicularfrom 11.Construction pointP islocatedatpointP islocatedatpointP Figure ofaperpendicular from 12.Construction pointP Figure ofaperpendicularfrom 10.Construction pointP 2 B . ThesegmentP 2 andAX 2 2 3 istheheighttosideBP . . 2 X 3 isthethirdheighttosideAB. 2 tothediameterABoritscontinuation. 1 3 tothediameterAB. tothediameterAB. 1 M 2 2 3 B isthethirdheight . PointX , whereBX isobtuse-angled. 3 isoutside 1 isthe A fascinating application of Steiner’s Theorem for Trapezium Australian Senior Mathematics Journal vol. 27 no. 2 19
, . . 5 2 B is is X 3 2 2 6 X and X , and 4 2 3 X , and a 1 4 BP , 2 , and draw 3 ), and draw . Point X (Figure 13). 4 4 2 and point X is the middle the is is connected AX 1 A and point X 1 is perpendicular to 5 (the construction is X 2 . Point B 4 4 , which stems from the , resulting in the point of in , resulting . The continuations of the 4 5 . 7 , so that trapezium X 6 with diameter AB (Figure 14). . Therefore, from property 3b, are, respectively, the points of the points are, respectively, 4 1 to diameter AB in the following or P X 4 with P 6 3 is perpendicular to diameter AB. 7 is connected with point P is connected with X are heights in the triangle ∆AX in the triangle are heights 3 and X 2 , P X 5 5 5 and N . This means that point that means This . 4 with point X is connected with point X and X 5 X 2 are the midpoints of its bases. Point bases. its of midpoints the are 2 . From property B . 7 and X 4 X N X 2 4 1 intersects the circle at point X intersects the circle and BX 1 X P 1 1 is the perpendicular to the diameter. , and point B 5 2 , on the circle (as the location of P at points X . Points M and AX 1 4 4 X 4 X M 3 , and the continuation of the line intersects the circle of the line intersects the circle , and the continuation , and point M 4 is located at point P is the third height that intersects the diameter at point M the diameter height that intersects is the third bisects angle bisects M 2 3 . A straight line is drawn through point X . A straight line . The line P 3 5 AX . The segment BX X 1 , and therefore, from property 2a, chord P , and therefore, from property 2a, Figure 13. Construction of a perpendicular from point P4 to the diameter AB. Construction 13. of a perpendicular from Figure 5 , is selected on the continuation of line AP , is selected on the is located at the point P is located at the X 1 . Ray . 4 is connected with X is connected with 4 X A X 2 are three heights in the triangle ∆ABX are three heights in the triangle 4 X 4 X 1 straight lines intersect at point X a perpendicular to point X and therefore X and therefore at the point X Steiner theorem for a trapezium, line P from it a perpendicular to the diameter to point X intersection is obtained, where points where obtained, is point, intersection of perpendiculars X diameter AB. cases): Draw perpendiculars X connected with point N of arc P connected with point X connected with point carried out as in case 4). In the same manner, choose a point, X carried out as in case 4). In the same manner, Description of the constructions: (The description is similar for all three Proof of the constructionProof : Proofs of the constructionsProofs P : Point manner: Choose a point, X Description of the constructionDescription of the : Point A Proof of the constructionthe of Proof segments cases, previous the in As : the continuation of the line intersects diameter AB at point X of the line intersects diameter the continuation we also obtain that they are the bisectors of the angles in the orthic triangle we also obtain that they are the with point X Point Case 4: P Cases 5, 6, and 7: P The segments intersect X ∆P X 20 Australian Senior Mathematics Journal vol. 27 no. 2 Stpel & Ben-Chaim An interestingquestionis:Whatextragivenconditionisneededtobeable Extension ofthetask F Then, construct a parallel line to lines tothegivensegmentCD(showninsection5.3above).Then,using locate thecentreofcirclewithonlyastraightedge? to theassignmentdescribedearlierinthispaper. the theoremaboutintersectionofthreealtitudesatriangleat chords case 1, from an arbitrary pointC case 1,fromanarbitrary on thesegmentABoritscontinuation),onemustapplySteinertheorem. intersects thegivendiameteratpointM diameter ofagivencircle(seeFigure16). perpendicular toalinethroughanexternalpoint),oneneedsonlyapply diameter. Steiner theoremfortrapeziumH Straightforward applicationoftheSteinertheorem: Anotheroptionisto Straightforward apply theSteinertheoremfortrapezium(seeFigure15).Withrespectto a diameterofthecircle(see Figure17). single point.However, for cases5–7(toconstructaperpendicularatpoint 1 F Through pointsH It is interesting to note that for the proofs of cases 1–4 (to construct a Choose twopointsonl Equivalently, given a circle and two parallel lines, it is possible to construct First, given asegment and itscentre point, it ispossibleto construct a 2 G 2 G H 1 , then, using the Steiner theorem, find themidpoints of thebases of 1 K Figure ofaperpendiculartothediameterABatpointP 14.Construction 1 andH Figure 15. Straightforward applicationoftheSteinertheoremFigure for 15.Straightforward constructing aperpendiculartothediameterABatpointP constructing 2 K 1 andH 2 (pointsL andtwopointsonk 2 onthecircle,itispossibletoconstructparallel X andN 1 , aperpendiculartoABisconstructed.It K 1 X 1 K 2 that passes through point 2 H ). ThechordthroughL , andthecircleatpointsX 2 , itispossibletobisecteachofthe anddrawthetrapezium 5 . P 5 . andN 5 according 1 andX isa 2 . A fascinating application of Steiner’s Theorem for Trapezium Australian Senior Mathematics Journal vol. 27 no. 2 21 , express the length of and b DF = FC. Using a α + β = 90°; AE = EB, Figure 17. Constructing a diameter of a given circle by given two parallel lines 17. Constructing a diameter of a given circle Figure ; Figure 16. Constructing a diameter of a given circle by given a segment CD and its midpoint M. by given 16. Constructingcircle a diameter of a given Figure Problem: Given is a trapezium with lengths of bases a and b, and where the Problem: Given is a trapezium with lengths of bases a and b, and where Therefore, to construct the centre of a given circle, we need two nonparallel Therefore, to construct the centre DC = b ; segments with their midpoints, or equivalently a parallelogram, or a triangle a parallelogram, or a or equivalently midpoints, with their segments sum of the angles of the lower base is 90°. That is, the following is given: AB = sum of the angles of the lower base is 90°. That is, the following is and its centre of gravity (the intersection point of its medians). In this case, and its centre of gravity (the intersection proven using this theorem. circle, which means that it is possible to construct with straightedge alone circle, which means that it is possible everywith compass and straightedge. construction that can be constructed with a straightedge alone. of a given circle by construction of a diameter of the circle. of a diameter trapezium in problem solving, we present another problem that can be easily trapezium in problem solving, we present another problem that can the trapezium (one midpoint is sufficient) and apply the previous construction previous construction and apply the is sufficient) (one midpoint the trapezium the segment connecting the midpoints of the bases (Figure 18). we are able to construct two diameters whose intersection is the centre of the we are able to construct two diameters In order to demonstrate the applicability of the Steiner theorem for the In order to demonstrate the applicability of the Steiner theorem Obviously, it is possible to create many more conditions for finding the centre it is possible to create many more conditions for Obviously, a Application of the task 22 Australian Senior Mathematics Journal vol. 27 no. 2 Stpel & Ben-Chaim ABCD withsidesAB=a Additional applications Another applicationoftheconstructionspresentedaboveistocreatesome with itsmidpoint(thediagonals ofthegivenparallelogram)toobtainanother using straightedgealone (seeFigure20). rhombus, constructusingstraightedgealonearectanglewithitssidesequalto trapezium, pointsG the trapezium,oruseoftrigonometrictools. to constructaparallelitthroughpointA the diagonalsofgivenrhombus (seeFigure19). the sameisappliedforsegmentACandpointsD Proof usingSteiner’stheorem: Extendthesidesuptotheirpointofintersection, either requires auxiliary constructionsandtheuseofalgebraicformulasfor either requiresauxiliary are equaltohalfthehypotenuse,andtherefore: and obtainaright-angledtriangle∆GAB.FromSteiner’s theoremfora again and again creates a nicetessellation pattern. Anotherpossibilityisto shapes thatcanbeusedeitherinartortessellation.Forexample,givena start withaparallelogramratherthanrhombus. GF and GE are medians to the hypotenuse in right-angled triangles, which Another exampleofapplicationisthefollowing task: Givenparallelogram Here too,usethemethod ofconstructingaparallellinetogivensegment Note: Otherproofsofthisproblemcanbegiven,buttheyarelongerand Now, itisobviousthat,giventhesegmentDBanditsmidpoint,possible Figure arectangle 19.Constructing withitssidesequaltothediagonalsofagivenrhombus Figure 18.ApplicationofSteiner’s theorem toexpress thelengthofFE. , E , andF andBC=b FE areononestraightline. = GE , constructlengthsm – GF = a , aswellthroughpointC 2 − b andB . Repeatingthisshape ⋅a andn ⋅b , m , n and ∈N A fascinating application of Steiner’s Theorem for Trapezium Australian Senior Mathematics Journal vol. 27 no. 2 23 corresponding sides a and b of a given parallelogram. Figure 20. Constructing the lengths of m and n times Figure In addition, several studies have pointed out that introducing historical historical introducing that pointed out have studies addition, several In This article is enhanced, the authors believe, by a brief history of geometric To this purpose, this article presents examples of geometric constructions to examples this purpose, this article presents To straightedge (without markings) alone. Indeed, Steiner claimed that everymarkings) straightedge (without sought.) Steiner’s theorem, a mathematical pearl that indicates the minimal sought.) Steiner’s significantly. significantly. shapes, as well as challenging thought. An important branch in this field An important branch in thought. challenging as well as shapes, ages are mentioned, with a description of how they coped with geometric ages are mentioned, with a description of how they coped with and impossible constructions based on rudimentaryand impossible constructions based methods construction Conclusion geometric construction requires the ability to differentiate between possible between differentiate to ability the requires construction geometric done using only a straightedge, given a circle and its centre (equivalent to one done using only a straightedge, given a circle and its centre (equivalent demythologises mathematics by showing that it is the creation of human demythologises mathematics by parallelogram, and then repeat this procedure as needed. In this case too, In this case as needed. repeat this procedure and then parallelogram, is geometric construction with restrictions on the types of tools used. Such is geometric construction with restrictions constructions, and by focusing specifically on constructions with aid of a constructions, and by focusing specifically on constructions with can be construction that can be conducted using a compass and straightedge conditions required for classical geometric constructions, is presented as a conditions required for classical geometric constructions, is presented construct a perpendicular on the diameter of a circle using only a straightedge. construct a perpendicular on the diameter of a circle using only a straightedge. of geometric construction through the history of mathematics enriches it one can adapt this work to art and tessellations. adapt this work one can construction problems, some of which involved proving that some specific construction problems, some of which involved proving that some more interesting and accessible (see for example Heiedi, 1996) and “it more interesting and accessible beings” (Marshal & Rich, 2000, p. 706). Hence, investigating the development beings” (Marshal & Rich, 2000, p. 706). Hence, investigating the development background to mathematical topics plays a vital role in making mathematics background to mathematical topics use of a compass). In addition, several mathematicians who lived in different use of a compass). In addition, several mathematicians who lived tool for the constructions. that, with their application, allow the performancethat, with their application, allow of more complicated ones. the good teacher. The geometric constructions topic is remarkable and can be The geometric constructions topic is remarkable and can the good teacher. theorems, and properties of geometrical used for implementing knowledge, Inspiring students to love and be excited by mathematics is one of the goals of Inspiring students to love and be (The centre was not given, and in some cases, the centre of the circle was also in some cases, was not given, and (The centre constructions were actually impossible, and which required the creation of new mathematical theories and proofs to finally prove them. Mention is made of the use of the dynamic geometry software as a helpful,
Stpel & Ben-Chaim powerful tool for teaching geometry, especially for enhancing the technical part of geometric constructions. The authors believe that the geometric constructions presented in this article can make a significant contribution to the education of both pre- and in- service mathematics teacher as well as to the high school geometry curriculum. Through such geometric constructions, including emphasising their place in the history of mathematics, the learner can expand their conception of mathematics and, hopefully, appreciate the fascination that is in mathematics.
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