THE DEVELOPMENT OF NAPOLEON’S THEOREM ON …

Bulletin of Mathematics ISSN Printed: 2087-5126; Online: 2355-8202

Vol. 08, No. 01 (2016), pp. 97–108. http://jurnal.bull-math.org

THE DEVELOPMENT OF NAPOLEON’STHEOREM ON QUADRILATERAL WITHCONGRUENCE AND TRIGONOMETRY

Chitra Valentika, Mashadi, Sri Gemawati

Abstract. This thesis discusses about Napoleon’s theorem on a quadrilateral thathas is two pairs of parallel side with two cases: (i) square built toward outsideand (ii) square built toward inside. The Napoleon’s theorem is proved by usingcongruence approach and trigonometric concepts. At the end of the discussion, theNapoleon’s theorem is developed by using the concept of intersecting parallel linesand using Geogebra applications.

1. INTRODUCTION

Remarkable math statements have been attributed to Napoleon Bona-parte (1769-1821) although his relation to the theorems and their proofs isquestioned in most of the sources available to our knowledge. Nevertheless,the mathematics flourished in post-revolutionary France and mathemati-cians were held in great esteem in the new Empire [10]. Napoleon’s theoremstates that if equilateral triangles are drawn on the sides of any triangle,either all outward, or all inward, the centroids of those equilateral trianglesare the vertices of an equilateral triangle [8]. Napoleons theorem on trianglecan be proved by elementary [9], and several articles discussing Napoleons

Received 12-07-2016, Accepted 20-07-2016.2010 Mathematics Subject Classification: 51F20, 51H10, 97G60Key words and Phrases: Napoleon’s theorem, Napoleon theorem on quadrilateral, square, con-gruence, trigonometry.

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Chitra Valentika, et. al. – The Development Of Napoleon’s Theorem ... 98

theorem proof with trigonometry [3][7]. There are two cases in the triangle,which is as follow.

Case 1. Napoleon’s theorem in first case describes an equilateral trianglecalled the outer Napoleon triangle constructed on each side of any trian-gle 4ABC toward the outside [2]. Let P,Q, and R is centroids of triangle4ABD, 4ACE, and 4BCF , the third of centroids form an equilateraltriangle called the external Napoleon triangle [1]. Illustrations shown in theFigure 1.

Case 2. Napoleon’s theorem in second case explain an equilateral triangleconstructed on each side of any triangle 4ABC toward the inside. LetX, Y, and Z is centroids of triangle 4ABD, 4ACE, and 4BCF the thirdof centroids form an equilateral triangle called the internal Napoleon triangle[8]. Illustrations shown in the Figure 2.

Figure 1: External Napoleon triangle

In this article discussed some of the result of proving theorems Napoleon’swith elementary geometry and trigonometry. Using charts excel, [4] statesthat the square be constructed on the each side of any quadrilateral, the fourcentroids of the square when connecting centers of square on the oppositeside then then both equal length and perpendicular. Then, according to [11]said that he tried several such as square, rhombus, rectangle, parallelogram,when constructed on each side of any quadrilateral, the four centroids of thesquare when connecting centers of square on the opposite side then the linesegments were clearly of equal length and perpendicular.


Figure 2: Internal Napoleon triangle

2. NAPOLEON’S THEOREM ON THE QUADRILATERAL

Napoleon’s theorem on the triangle is the development of theoremNapoleon on the triangle. Several experiments conducted found for square,rhombuses, rectangles, and parallelograms forming a square, for an isoscelestrapezium forming kites, and for any quadrilateral forming any quadrilat-eral. There are two cases of Napoleon’s theorem on the quadrilateral shapedparallelogram is as follows.

Case 1. In case 1, Let ABCD denote any parallelogram, square construc-tion on each side of the parallelogram outwards, shown in the Figure 3.

Teorema 2.1 If M,N,O, and P is the centroid of each square ABHG,square ADEF , square CDKL, and square BCIJ which constructed on eachside of the parallelogram outwards. The fourth centroids of the square con-nected so form a square MNOP .

Proof 1. By4GAD and4BAF , obtained AG = AB,∠GAD = ∠FAB,AD =AF , so 4GAD ≈ 4BAF [6]. Perform 4GQT and 4BAT in Figure 4∠TGQ = ∠TBA and ∠GTQ = ∠BTA, then ∠GQT = ∠BAT = 90o. Then∠GQT = ∠QSR = ∠MRN = 90o, similarly ∠MRN = ∠QSR = 90o.Perform 4MV R and 4NWR clear MV = WR, ∠MV R = ∠NWR,V R =NW , therefore 4MV R ≈ 4RWN . Since 4MV R ≈ 4RWN , then MR =


Figure 3: Square construction on each side of the parallelogram outwards

Figure 4: Squares which constructed on each side of the parallelogram out-wards


RN , similarly for OR = PR, therefore4MRP ≈ 4NRO. Since4MRP ≈4NRO, then PM = ON similarly MN = OP . it is evident that OM =NP and OM⊥NP .

Proof 2. Let AB = CD = a, side AC = BD = b, then AM = MB = OC =OD = 1

2a√

2AM = MB = OC = OD = 1

2b√

2

Figure 5: Triangle construction

Using the cosine rule [5] in 4MAN applyMN2 = 1

2a2 + 12b2 − 2.12a

√2.12b

√2. cos ∠MAN

MN2 = 12a2 + 1

2b2a.b. cos(270o − ∠BAD)MN2 = 1

2a2 + 12b2a.b.(cos 270o. cos angleBAD + sin 270o. sin∠BAD)

MN2 =12a2 +

12b2 + a.b. sin∠BAD (1)

And in 4ABD apply

sin∠BAD =2.L4ABD

a.b(2)

Substituting equation 2 to 1 in order to obtain


MN2 = 12a2 + 1

2b2 + a.b.2.L4ABD

a.b

MN =√

12a2 + 1

2b2 + 2.L4ABD

MN =

√12a2 +

12b2 + A.parallelogram ABCD (3)

Subsequently in 4NDO obtained

NO2 = 12a2 + 1

2b2 − 2.12a√

2.12b√

2. cos ∠NDONO2 = 1

2a2 + 12b2 − a.b. cos(90o + ∠DAC)

NO2 = 12a2 + 1

2b2 − a.b.(cos 90o. cos ∠ADC − sin 90o. sin∠ADC)

NO2 =12a2 +

12b2 + a.b. sin∠ADC (4)

Figure 6: Square construction on each side of the parallelogram inwards

Teorema 2.2 Since t =A.parallelogram ABCD

band sin∠ADC = t

a thenapply

sin∠ADC =A.parallelogram ABCD

a.b(5)


Substituting equation 5 to 4 in order to obtain

NO2 = 12a2 + 1

2b2 + a.b.A.parallelogram ABCD

a.b

NO =

√12a2 +

12b2 + A.parallelogram ABCD (6)

Since MN = NO similarly MN = OP and NO = MP , therefore MN =OP = NO = MP , and the second diagonal intersect perpendicularly [8].This completes the proof of theorem 1.

Case 2. In case 2, let ABCD denote any parallelogram, square constructionon each side of the parallelogram inwards, shown in the Figure 6.

If M, N, O, and P is the centroid of each square ABHG, square ADEF ,square CDKL, and square BCIJ which constructed on each side of the par-allelogram inwards. The fourth centroids of the square connected so form asquare MNOP .

Figure 7: Squares which constructed on each side of the parallelogram in-wards

Proof. Notice that line GD and BF extended then intersected at the pointS and the point V , shown in the Figure 7. Moreover on 4BUF and


4TSB,∠TBS = ∠BFU, ∠FBU = ∠STB, therefore 4BUF ≈ 4TSB,then ∠FUB = ∠TSB = 90o. Since V M ′ = O′M,∠V M ′R = ∠MO′R, V R =RM therefore4V M ′R ≈ 4MRO′, then M ′R = O′R, similarly for4P ′QR ≈4N ′NR, then N ′R = P ′R. Let P ′N ′ and O′M ′ are diagonal on the squareM ′N ′O′P ′ so dividing the same diagonal length and intersect perpendicu-larly at the point R. Then proved O′M ′N ′P ′ is square.

3. THE DEVELOPMENT OF NAPOLEON’S THEOREM ONQUADRILATERAL

Napoleon’s theorem on quadrilateral developed based Napoleon’s the-orem on quadrilateral for case square built leads outward.

Teorema 3.3 If Q,R, S, and T is the midpoint of the line FG, EL,KJ ,and HI, then QRST is a square.

Figure 8: Napoleon’s theorem on quadrilateral

Proof. Since FN = PJ, ∠QFN = ∠PJS, and FQ = SJ , therefore4FNQ ≈4PJS then QN = PS. Similarly PI = NE, ∠NER = ∠PIT , and IT =


ER, therefore 4NER ≈ 4PIT then NR = TP . Since QN = PS, NR =TP and ∠QNR = ∠SRT , therefore 4QNR∠4SPT , then QR = TS. Sim-ilarly 4TMQ and 4SOR, TM = OR,∠TMQ = ∠SOR, and OM = OS,therefore 4TMQ ≈ 4SOR then TQ = SR. Since ∠QUR = ∠TUS, and∠QUT = ∠RUS therefore ∠QUT = ∠RUS = ∠QUR = ∠TUS = 90o,then proved QRST is square.

The following corollary proves statements 1, 2, and 3 from theorem 3.

Corollary 3.1 On the square MNOP and TQRS formed parallel linesPQ//SN and MR//TO, then formed a square V WZU , and if formedparallel lines MS//QO and TN//PR, then formed a square A1B1C1D1.Illustrations in Figure 9.

Figure 9: Parallel lines PQ//SN and MR//TO

Corollary 3.2 On the square V WZU and A1B1C1D1 formed parallel linesUA1//V C1 and ZD1//WB1, then formed a square K1N1M1H1, and ifformed parallel lines V B1//UD1 and ZD1//WB1, then formed a squareO1P1Q1R1. Illustrations in Figure 10.

Corollary 3.3 On the Figure K1N1M1H1 and O1P1Q1R1 formed paral-lel lines K1P1//M1R1 and K1Q1//M1O1, then formed a square E1F1I1G1,and if formed parallel lines N1R1//M1O1 and K1Q1//M1O1, then formed asquare J1L1T1S1. Illustrations in Figure 11.


Figure 10: Parallel lines V B1//UD1 and ZD1//WB1

Figure 11: Parallel lines N1R1//M1O1 and K1Q1//M1O1


4. CONCLUSION

Napoleon’s theorem at quadrilateral only to the quadrilateral that hastwo pairs of parallel side, such as square, rectangle, rhombus, parallelogram.Then development with two cases: (i) square built toward outside and (ii)square built toward inside then the fourth centroids of the square connectedso form a square.

REFERENCES

1. B.J.Mc. Cartin, Mysteries of the Equilateral Triangle, Hikari Ltd, 978-954-91999-5-6(2010), 36-37.

2. G.A. Venema, Exploring Advanced Euclidean Geometry with Geometer’sSketchpad, Grand Rapids, Michigan 49546 (2009), 84-85.

3. J.A.H. Abed, A Proof of Napoleon’s Theorem, The General Science Jour-nal, 1(2009), 2-4.

4. J. Baker, Napoleon’s Theorem and Beyond, Spread Sheets in Educations(eJSiE), 1[4], Bond University’s Repository, 2005.

5. M. Corral, Trigonometry, Schoolcraft collage, GNU Free DocumentationLicense, Version 1.3, Livonia, Michigan, 2009.

6. Mashadi, Buku Ajar Geometri, PUSBANGDIK UNRI, Pekanbaru, 2012.7. N. A.A. Jariah, Pembuktian Teorema Napoleon dengan Pendekatan Trigonometri,

[http://www.academia.edu/12025134/ Isi NOVIKA ANDRIANI AJ 06121008018],accessed 6 October 2015

8. P. Bredehoft, Special Cases of Napoleon Triangles, Master of Science,University of Central Missouri, 2014.

9. P. Lafleur, Napoleons Theorem, Expository paper, [http://www.Scimath.unl.edu/MIM/files/MATEexamFiles], accessed 24 November 2015.

10. V. Georgiev and O. Mushkarov, Around Napoleon’s Thorem, LifelongLearning Progamme, 510028, Dyna Mat, 2010.

11. Y. Nishiyama, Beatiful Geometry As Van Aubel’s Theorem, 533-8533,Univ. Osaka, Dept. Business information, 2010.

Chitra Valentika: Magister Student, Department of Mathematics, Faculty ofMathematics and Natural Scinces Uneversity of Riau, Bina Widya Campus, Pekan-baru 28293, Indonesia.E-mail: [email protected]


Mashadi: Analysis group, Department of Mathematics, Faculty of Mathematicsand Natural Scinces Uneversity of Riau, Bina Widya Campus, Pekanbaru 28293,Indonesia.E-mail: [email protected]

Sri Gemawati: Analysis group, Department of Mathematics, Faculty of Mathe-matics and Natural Scinces Uneversity of Riau, Bina Widya Campus, Pekanbaru28293, Indonesia.E-mail: [email protected]

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THE DEVELOPMENT OF NAPOLEON’S THEOREM ON …