Multiple Proofs for a Geometric Problem

Transcription

1 Multiple Proofs for a Geometri Problem Lee Tuo Yeong.Tan Eng Guan Toh Tin Lam. Dong engming Mathematis and Mathematis Eduation National Institute of Eduation Nanyang Tehnologial University Singapore Corresponding author: tylee@nie.edu.sg

2 VOLUIIU ~ JUO.( 20m of a etri roblem Multiple Proofs for a Geometri Problem Introdution The following is a typial plane geometry problem: ACD is a quadrilateral with A = CD. Let E and be the midpoints of AD and C respetively. Lines CD and E produed meet at point G and lines A and E produed meet at point H, as shown in igure 1. Prove that L.H = L. CG. igure 1 While the problem is not very diffiult to solve, it will be instrutive to eluidate from students as many different proofs as possible. This has the dual purpose of enouraging reativity and raising the important awareness that quite often, there are many ways (sometimes using different areas of mathematis) to solve a partiular mathematis problem. 44 Mathematial Medley

3 YOLUIM lo 00.1, JUM 20m root'- on1et ~. rob n To these ends, this paper presents ten different proofs of the problem, five omplete and five skethes. These proofs involve the following areas in mathematis. (i) Geometry (ii) Trigonometry (iii) Vetors (iv) Analyti geometry Construting New Line Segments of Extending Old Ones The method of onstruting new line segments or extending old ones is very ommon in geometry proofs. One must not be afraid to 'dirty' the diagram with new lines and new points to get a better view of the problem. Line segments drawn (and their extensions) are usually parallel or perpendiular to existing lines or join existing points. Points are also suitably hosen (midpoints, points of intersetion, et.). Proof 1 igure 2 Join and D, and let I be the mid-point of D. Then join I to both E and, as shown in igure 2. Sine AE =ED and I = ID, we have EI = Yz A and EI II A. Similarly, we have I = Yz CD and I II CD. Sine A = CD, we have EI =l. Thus LIE= LIE. Sine EI I I A, we have LIE= LH. Sine I I I CD, we have LIE= L CG. Hene LH= LCG. Mathematial Medley 45

4 VOLUIM ~0 RO.I, )UIU 20m M tiple ofs ag metri Problem Note: Here we have used the fat that the line joining the mid-points of two sides of triangle is always parallel to and half the length of the third side. This is a very useful fa in solving many geometry problems. Proof 2 igure 3 Extend E to point Q suh that EQ =E. Join point Q to both points C and D. Let CG and AQ meet at point P. E=EQ} _ =C =>E II QC=>LCG-a. AE=DE } { LAE = LDEQ => ME = WEQ => LAE = LD'!_E => A II DQ M=~ ~-~ A=DQ} A = DC => DQ = DC =>a = p. LH = LH -LH} E II CQ => LH = LCQ => LH = 180 -LCQ-LQC- LDQE = p. A II QD => LHE = LDQE a =P } a=lcg =>LCG=LH. P=LH 46 M>thematial Medley

5 YOLUIM lo 00.1, JUM 20m Multi pi Proof a eometri roblem Some Trigonometry- Using Sine Rule The next proof involves the appliation of the sine rule. Proof 3 Applying the sine rule to tillh gives sinlh sinlh = H Applying the sine rule to J..CG gives sinlcg sinlcg = C GC Sine LH + L CG = 180, we have sin LH = sin L CG. Sine = C, it follows from (1), (2) and (3) that sinlh GC ---= sinlcg H In the same way, we an show that sinlh DG = sinlcg AH Thus it follows from (4) and (5) that GC DG -=- H AH Sine A =DC, it is easy to show by (6) that H = GC. Then, by (4), we have sinlh = sinlcg. (1) (2) (3) (4) (5) (6) (7) Sine LH + LH < 180 => LH + LCG + LCG < 180 => LH + LCG < 180, it follows from (7) that LH=LCG.

6 vownu ~o no.1, JUM 2om ML1Itiple oafs o a Geometri Problem Vetors and Analyti Geometry Vetors are a natural way of representing line segments in spae. One thus represented, the manipulation of the vetors may proeed without muh referene to the diagram, whih is espeially useful in three-dimensional geometry. The fmal two proofs involve vetors - the last proof having an analyti geometry flavour by framing the diagram within a oordinate system. Proof 4 H igure E = EA + A + = t DA + A + t C E = ED + DC + C = tad + DC + t C. Hene E = t(a +DC). Sine ( A + OC) o (A- OC) = A 2 - DC 2 = 0, we have --- E o (A- DC)= 0. Hene, we have E o A = E o DC Ex Ax oslh = Ex DCx oslcg oslh = oslcg. Sine the two angles are between oo and 180, we have LH = LCG. 48 Noothemotial Medley

7 YOLUIJI.( ~0 00.1, JUO.{ 20m 1\ L ltiph Proof eometr m Proof 5 X igure 5 Let be the origin of the oordinate system, and and C be the points (-a, 0) and (a, 0) respetively. Let CD = A = b. Then the oordinates of points A and D are (-a+bosa, bsina) and (a+bos~, bsi~) for some a and~ Sine E is the midpoint of AD, the oordinates of E are (tb(osa +os~ ),tb(sina +sin~)). Thus the vetors A, E and CD are A = (bosa,bsina), - CD = ( b os ~, b sin ~ ). E=(tb(osa +os~),tb(sina +sin~)), Now, A o E= (bosa x tb(osa +os~))+ (bsina x tb(sina +sin~))= tb 2 (1 + os(a- ~ )). Similarly, CD o E= tb 2 (1 + os(a -13 )) Sine A =CD, A o E= Ax Ex oslh, CD o E= CDx Ex oslcg, we have oslh = oslcg. Sine the two angles are between 0 and 180, we have LH = LCG. Note: When other approahes seem to fail, using oordinate geometry approah (as above) usually works. However, it takes quite a bit of skill to frame the diagram using a onvenient oordinate system and setting the orret number of variables. Mathematial Medley 49

8 YOLUIIU ;ao RO.I, JUIU 20m Multiple Proofs for a Geometri Problem Another ive Proofs (Skethes Only) In this setion we provide another five solutions, albeit skethes only. The reader is enouraged to omplete the proofs. Proof 6 {Sketh) Statement: Extend H to point W suh that W = H. Join C and W. Join A and and extend A to point Q on CW. Draw line QP parallel to DA, where P is a point on W. Q igure 6 50 Noathematiol Medley

9 YOLUIM lo JUM 200) Multiple Proofs f a Geometri I:Jro blem Proof 7 (Sketh) Statement: Draw lines AP and DQ suh that APIIEIIDQ, where P and Q are points on C. igure 7 Proof 8 (Sketh) Statement: Draw lines AP and DQ parallel to C, where P and Q are two points onh. igure 8 Mathematial Medley 51

10 volum-e: ~o no.1, JUn-e: 2om a elri Problem Proof 9 (Sketh) Statement: Draw lines P and CQ parallel to E, where P and Q are on the extension of the line AD. igure 9 Proof 1 0 (Sketh) H Statement: Draw line AP parallel to GC, where P is a point on the extension of G. igure 10 p 52 Nathematial Medley

11 VOLUm+ ~0 00.1, JUR+ 20m ro rnet b n Conlusion The ten proofs should onvine mathematis problem solvers that there are often many solutions to a single problem. In fat, in a book by Elisha Loomis [1], there are 365 more or less distint proofs of Pythagoras' Theorem! We enourage the reader to attempt a nie geometry (or some other mathematis) problem using different approahes and we trust that the reader will then gain greater insight into mathematis from the approahes used. Referenes 1. Loomis, E., The Pythagorean Proposition, National Counil of Teahers of Mathematis, (1968). MJthematial Medley 53