THE DEVELOPMENT OF NAPOLEON’S THEOREM ON QUADRILATERAL WITH CONGRUENCE AND TRIGONOMETRY

  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’S

THEOREM ON QUADRILATERAL WITH

  

CONGRUENCE AND TRIGONOMETRY

_Abstract.

Chitra Valentika, Mashadi, Sri Gemawati

This thesis discusses about Napoleon’s theorem on a quadrilateral that

has is two pairs of parallel side with two cases: (i) square built toward outside

and (ii) square built toward inside. The Napoleon’s theorem is proved by using

congruence approach and trigonometric concepts. At the end of the discussion, the

Napoleon’s theorem is developed by using the concept of intersecting parallel lines

and 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 is questioned 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 theorem states that if equilateral triangles are drawn on the sides of any triangle, either all outward, or all inward, the centroids of those equilateral triangles are the vertices of an equilateral triangle [8]. Napoleons theorem on triangle can be proved by elementary [9], and several articles discussing Napoleons Received 12-07-2016, Accepted 20-07-2016.

  2010 Mathematics Subject Classification: 51F20, 51H10, 97G60

Key words and Phrases: Napoleon’s theorem, Napoleon theorem on quadrilateral, square, con-

gruence, trigonometry.

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

  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 triangle called the outer Napoleon triangle constructed on each side of any trian- gle △ABC toward the outside [2]. Let P, Q, and R is centroids of triangle △ABD, △ACE, and △BCF , the third of centroids form an equilateral triangle called the external Napoleon triangle [1]. Illustrations shown in the Figure 1. Case 2

  . Napoleon’s theorem in second case explain an equilateral triangle constructed on each side of any triangle △ABC toward the inside. Let X, Y, and Z is centroids of triangle △ABD, △ACE, and △BCF the third of 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’s with elementary geometry and trigonometry. Using charts excel, [4] states that the square be constructed on the each side of any quadrilateral, the four centroids of the square when connecting centers of square on the opposite side 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 the square when connecting centers of square on the opposite side then the line segments were clearly of equal length and perpendicular. Chitra Valentika, et. al. – The Development Of Napoleon’s Theorem ...

  Figure 2: Internal Napoleon triangle

  2. NAPOLEON’S THEOREM ON THE QUADRILATERAL Napoleon’s theorem on the triangle is the development of theorem

  Napoleon on the triangle. Several experiments conducted found for square, rhombuses, rectangles, and parallelograms forming a square, for an isosceles trapezium forming kites, and for any quadrilateral forming any quadrilat- eral. There are two cases of Napoleon’s theorem on the quadrilateral shaped parallelogram 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 each side of the parallelogram outwards. The fourth centroids of the square con- nected so form a square M N OP . Proof 1. By △GAD and △BAF , obtained AG = AB, ∠GAD = ∠F AB, AD = AF , so △GAD ≈ △BAF [6]. Perform △GQT and △BAT in Figure 4 _o

  . Then ∠T GQ = ∠T BA and ∠GT Q = ∠BT A, then ∠GQT = ∠BAT = 90 _{o o} ∠GQT = ∠QSR = ∠M RN = 90 .

  , similarly ∠MRN = ∠QSR = 90 Perform △MV R and △NW R clear MV = W R, ∠MV R = ∠NW R, V R = N W , therefore △MV R ≈ △RW N. Since △MV R ≈ △RW N, then MR = Chitra Valentika, et. al. – The Development Of Napoleon’s Theorem ...

  Figure 3: Square construction on each side of the parallelogram outwards Figure 4: Squares which constructed on each side of the parallelogram out- wards

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

  RN , similarly for OR = P R, therefore △MRP ≈ △NRO. Since △MRP ≈ △NRO, then P M = ON similarly MN = OP . it is evident that OM = N P and OM ⊥NP .

  Proof 2. Let AB = CD = a, side AC = BD = b, then AM = M B = OC = √

  1 OD = a

  2

  2

  √

  1 AM = M B = OC = OD = b

  2

  2 Figure 5: Triangle construction

  Using the cosine rule [5] in △MAN apply √ √

  2

  1

  2

  1

  2

  1

1 M N = a + b a

  2. b 2. cos ∠MAN − 2.

  2

  2

  2

  2 _o

  2

  1

  2

  1

  2

• M N = a b a.b. cos(270

  − ∠BAD)

  2

  2 _{o o}

  1

  1

  2

  2

  2

• M N = a b a.b.(cos 270 . cos angleBAD + sin 270

  . sin ∠BAD)

  2

  2

  1

  1

  2

  2

  

2

M N + = a b (1)

• a.b. sin ∠BAD

  2

2 And in △ABD apply

  2.L△ABD (2) sin ∠BAD = a.b

  Substituting equation 2 to 1 in order to obtain

2.L△ABD

  a

  2 + a.b.

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

  M N

  2

  =

  1

  2

• 1
• 1
• 2.L△ABD M N = r 1
• 1
• A.parallelogram ABCD (3) Subsequently in △NDO obtained N O

  2

• 1

  2

  a

  2

  2

  b

  2

  − a.b. cos(90 ^o + ∠DAC) N O

  2

  =

  1

  2

  a

  2

  b

  1

  2

  − a.b.(cos 90 ^o . cos ∠ADC − sin 90 ^o

  . sin ∠ADC) N O

  2

  =

  1

  2 a

  2

  2 b

  

2

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

  Teorema 2.2 Since t =

  A.parallelogram ABCD b and sin ∠ADC = ^{t a} then apply sin ∠ADC =

  A.parallelogram ABCD (5)

  2

  2

  =

  2

  b

  a.b M N = ^q

  1

  2

  a

  2

  2

  b

  2

  2 a

  2

  2 b

  2

  2

  =

  1

  2

  a

  2

  2

  b

  2 − 2.

  1

  2

  1

  2

  b √

  2. cos ∠NDO N O

  a √ 2.

• 1
• 1
• 1
• a.b. sin ∠ADC

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

  Substituting equation 5 to 4 in order to obtain A.parallelogram ABCD

  2

  1

  2

  1

2 N O = a b + + a.b.

  2

  2

  a.b r 1

  1

  2

2 N O = + a b + A.parallelogram ABCD (6)

  2

  2 Since M N = N O similarly M N = OP and N O = M P , therefore M N = OP = N O = M P , 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 construction on 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 a square M N OP .

  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 point S and the point V , shown in the Figure 7.

  Moreover on △BUF and Chitra Valentika, et. al. – The Development Of Napoleon’s Theorem ...

  △T SB, ∠T BS = ∠BF U, ∠F BU = ∠ST B, therefore △BUF ≈ △T SB, _o

  ′ ′ ′ ′

  . Since V M = O R, V R = then ∠F UB = ∠T SB = 90 M, ∠V M R = ∠MO

  ′ ′ ′ ′ ′

  , then M R = O RM therefore △V M R ≈ △MRO R, similarly for △P QR ≈

  ′ ′ ′ ′ ′ ′ ′

  N R, then N R = P R. Let P N and O M are diagonal on the square △N

  ′ ′ ′ ′

  M 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 ON QUADRILATERAL

  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 F G, EL, KJ, and HI, then QRST is a square.

  Figure 8: Napoleon’s theorem on quadrilateral Proof. Since F N = P J, ∠QF N = ∠P JS, and F Q = SJ, therefore △F NQ ≈ △P JS then QN = P S. Similarly P I = NE, ∠NER = ∠P IT , and IT = Chitra Valentika, et. al. – The Development Of Napoleon’s Theorem ...

  ER, therefore △NER ≈ △P IT then NR = T P . Since QN = P S, NR = T P and ∠QNR = ∠SRT , therefore △QNR∠△SP T , then QR = T S. Sim- ilarly △T MQ and △SOR, T M = OR, ∠T MQ = ∠SOR, and OM = OS, therefore △T MQ ≈ △SOR then T Q = SR. Since ∠QUR = ∠T US, and _o ∠QU T = ∠RU S therefore ∠QU T = ∠RU S = ∠QU R = ∠T U S = 90 , then proved QRST is square.

  The following corollary proves statements 1, 2, and 3 from theorem 3. Corollary 3.1

  On the square M N OP and T QRS formed parallel lines P Q//SN and M R//T O, then formed a square V W ZU , and if formed parallel lines M S//QO and T N//P R, then formed a square A B C D .

  1

  1

  1

  1 Illustrations in Figure 9.

  Figure 9: Parallel lines P Q//SN and M R//T O Corollary 3.2

  On the square V W ZU and A B C D formed parallel lines

  1

  1

  1

  1 U A //V C and ZD //W B , then formed a square K N M H , and if

  1

  1

  1

  1

  1

  1

  1

  1

  formed parallel lines V B //U D and ZD //W B , then formed a square

  1

  1

  1

  1 O

  1 P

  1 Q

  1 R 1 . Illustrations in Figure 10.

  Corollary 3.3 On the Figure K N M H and O P Q R formed paral-

  1

  

1

  1

  1

  1

  1

  1

  1

  lel lines K P //M R and K Q //M O , then formed a square E F

  I G ,

  1

  1

  1

  1

  1

  1

  

1

  1

  1

  1

  1

  1

  and if formed parallel lines N R //M O and K Q //M O , then formed a

  1

  1

  

1

  1

  1

  1

  1

  1 square J L T S . Illustrations in Figure 11.

  1

  1

  1

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

  Figure 10: Parallel lines V B //U D and ZD //W B

  1

  1

  1

  1 Figure 11: Parallel lines N R //M O and K Q //M O

  1

  1

  1

  1

  1

  1

  1

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

  4. CONCLUSION Napoleon’s theorem at quadrilateral only to the quadrilateral that has two 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 connected so form a square.

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7. N. A.A. Jariah, Pembuktian Teorema Napoleon dengan Pendekatan Trigonometri,

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MIM/files/MATEexamFiles], accessed 24 November 2015.

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

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  Chitra Valentika : Magister Student, Department of Mathematics, Faculty of Mathematics and Natural Scinces Uneversity of Riau, Bina Widya Campus, Pekan- baru 28293, Indonesia.

  E-mail: chitra.valentika@yahoo.com

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

  Mashadi : Analysis group, Department of Mathematics, Faculty of Mathematics

and Natural Scinces Uneversity of Riau, Bina Widya Campus, Pekanbaru 28293,

  Indonesia.

  E-mail: mashadi.mat@gmail.com

  Sri Gemawati : Analysis group, Department of Mathematics, Faculty of Mathe-

matics and Natural Scinces Uneversity of Riau, Bina Widya Campus, Pekanbaru

  28293, Indonesia.

  E-mail: gemawati.sri@gmail.com