International Journal of Mathematics Research. ISSN 09765840 Volume 4, Number 4 (2012), pp. 455461 © International Research Publication House http://www.irphouse.com
Eritrea’s Theorem Sied Mohammed Ali Orotta School of Medicine, P.O. Box 10549, Asmara, Eritrea, North East Africa Email: allabolosto@gmail.com
Abstract
This paper deals with a mathematical theorem, aα + cγ = bβ that could be applicable in many areas, e.g. mathematics, physics, engineering, astronomy, navigation, survey...etc. Triangles are applied in many practical measurements (e.g. survey). Therefore, it is important to make calculations from limited data about triangles. In showing (displaying) the proof for this theorem, Pythagoras theorem and Extensions of Pythagoras theorem were used. Likewise, triangles were classified generally as Acute angled triangle, Right angled triangle and Obtuse angled triangle. The medians and heights were taken as key instruments for arrival to the conclusion. The term ‘’sma’’ was introduced to make concepts easy and calculations simple. To make calculations easy the distance on each side (α, γ or β) between the intersection point of median and its corresponding base, and the altitude and its corresponding base is termed ‘’sma’’. Sma on side a is α, on side b is β and on side c is γ. The relation this research paper reveals is used on any triangle .Therefore; it holds a crucial importance to problems of solving triangles. This relation states that the product of the middle side and its sma is equal to the sum of the products of the remaining sides and their respective smas (the sum of the products of the length of the longest side and its sma and the length of the shortest side and its sma).
Keywords: Sma, Height, Median, Theorem, Triangle, Acute, Right, Obtuse, Proven
Introduction
The regularity in the shapes of various objects of nature aroused the geometrical sense in man ever since the prehistoric times. Though gradual processes of refinement and obstruction the ancient Greeks and Egyptians obtained various results in respect of the 456 Sied Mohammed Ali geometrical objects and applied their knowledge with profit in many activities (DAS, G.et al, 2000 ).Triangles are involved in many practical measurements (e.g. survey). So it is important to be able to make calculations from limited data about a triangle.
Although a triangle has three sides and three angles, it is not necessary to know all of these in order to define a particular triangle. If enough information about a triangle is known, the remaining sides and angles can be calculated by using a suitable formula. This is called ‘’solving the triangle’’. The two relationships that are used most frequently are the sine rule and the cosine rule (BOSTOCK.L and CHANDLER,S.,2000).In many branches of science like physics, mathematics, astronomy, engineering, navigation, survey etc measuring distances, length of objects, velocities ….etc can be worked out as functions to the problems of solving triangles.
For this purpose, many theorems were discovered. Some among the many could be mentioned as follows; Pythagoras theorem, Extensions of Pythagoras theorem, Double angle triangle theorem, Triple angle triangle theorem, Apollonius theorem and mollweid’s equation. In continuation to this, this paper discusses and gives the proof for a mathematical theorem (aα+cγ=bβ) that can be manipulated as are the preceding theorems.
Methodology
In general, triangles were classified in to three types based on the angles they contained. Here, in this classification, the three types of triangles are; Acute angled triangle, Right angled triangle, and the Obtuse angled triangle. The proof for this theorem is suitably and separately shown for each type of triangle and eventually, concluded for all of them i.e., for any triangle.
To prove the theorem, Pythagoras theorem and Extension of Pythagoras theorem were used, likewise, the term ‘’sma’’ for the segment (α, γ or β) of each side of the triangle was introduced. In some cases the three medians and the three altitudes were used up as key instruments for the proof. To make calculations easy the distance on each side (α,γ or β) between the intersection point of median and its corresponding base, and the altitude and its corresponding base is termed ‘’sma’’. For clarity purpose, the extension of Pythagoras theorem is stated here.
Figure 1
Eritrea’s Theorem
In ΔABC, ∠ A > 900 In ΔABC, ∠ A > 900 BC = a, AB = c, AC = b Then, a2 = c2 + b2 +2cx
Result Proof The proof for this theorem is given for each type of triangle (Acute, Obtuse, and Rightangled triangle) and eventually concluded for all types of triangles (i.e. for anytriangle). It is as follows; Acute Angled Triangle Labels AB=c AD =h AC=b AM =m BC=a DM = α
Figure 2
Given c ≥ b ≥ a ABC is an acute  angled triangle. M is the midpoint of BC AM is a median. AD is ⊥ to BC ∠D = 900 α is a sma of BC. (AD is ⊥ to BC& AM is the median) To prove the relation aα +cγ = bβ, see below h2 =b2 – (a/2 – α)2 (Pythagoras Theo.) 458 Sied Mohammed Ali h2 = b2 – a2/4 + aα – α2 e q. 1 In ΔABD, ∠ D = 900, therefore; h2 =c2 – (a/2 + α) 2 (Pythagoras Theo.) h2 = c2 – a2/4 – aα – α2 e q. 2 By subtracting eq.1 from eq.2 we can calculate sma of BC. Therefore; α = (C2 – b2)/2a Similarly, β = (c2 – a2)/2b and γ = (b2 – a2)/2c.Where they are smas of AC & AB respectively. From this it follows that; 2aα +2cγ +2bβ = 2a (C2– b2)/2a + 2c (b2 – a2)/2c +2b (c2– a2)/2b 2aα +2cγ +2bβ = 2C2 –2a2, where (c2 – a2) = 2bβ (from the above i.e. β = (c2 – a2)/2b), therefore; 2aα +2cγ +2bβ =4 bβ aα +cγ = bβ, hence it is proven. Obtuse Angled Triangle
Figure 3
Labels AB =c AZ =MEDIAN1 BD =h2 EC =n BC =a BN =MEDIAN2 CS =h3 DC =x CA =b CM =MEDIAN3 AE =h1 SM =γ Given c ≥ b≥ a ΔABC is an Obtuse Angled triangle, ∠ C > 900. BE is ⊥ to AE DB ⊥ to AD
CS ⊥ to AB In ΔABC, ∠ C > 900, therefore, from the extension of Pyth.Theo; C2 =a2 +b2 +2bx where ‘x’ is the extension made on AC i.e. x = CD then, X = (c2 – a2 – b2)/2b, but sma (β) of AC is (x+ b/2) (BD is ⊥ to AD produced & BN is the median of AC) Therefore; β = (X+ b/2) = (c2 – a2 – b2)/2b + b/2 β = (C2– a2)/2b sma ( α) of side BC is (n+ a/2). See below; In Δ AEB, ∠ E = 900, therefore; h2 1= c2 – (n+a)2 (Pyth.theo.) h2 1 =c2 – n2– 2an – a2 e.q.1 And In Δ AEC, ∠ E = 900 therefore, h2 1 =b2 – n2 e.q. 2 c2 – n2 –2an – a2 = b2 – n2 combining both e.q 1 and e.q. 2 (c2– b2 – a2) / 2a = n α = (n+a/2) (AE is ⊥ to BE produced & AZ is the median of BC) So it is calculated as follows; α = (c2 – b2 – a2) / 2a + a/2 α = (c2 – b2 )/2a Likewise, sma ( γ) of side AB is SM ( CS is ⊥ to AB & CM is the median of AB) In ΔCSA, ∠ S =900, therefore, h2 3 = b2– (c/2+γ) 2 (Pyth.theo.) h2 3 = b2– c2/4 – cγ – γ2 e.q.1 In ΔCSB, ∠ S =900, therefore; h2 3 = a2 – (c/2 – γ)2 (Pyth.theo.) h2 3 =a2 – c2/4 +cγ – γ2 e.q 2 γ = (b2 – a2)/2c combining e.q.1and e.q. 2 Similarly, as it was done for the acute angled triangle; 2aα +2cγ +2bβ = 2a (C2 – b2)/2a + 2c (b2 – a2 )/2c +2b (c2 – a2)/2b 2aα +2cγ +2bβ = 2C2 – 2a2, where (c2 – a2) = 2bβ (from the above), Therefore; 2aα +2cγ +2bβ =4 bβ aα + cγ = bβ, hence it is proven.
Right Angled Triangle Labels AB =c CD =h BN =MEDIAN1 BC =a CS =m AM =MEDIAN2 AC =b DS =γ CS =MEDIAN3
Figure 4
Given c > b ≥ a ΔABC is Right angle triangle. ∠C = 900 CD is ⊥ to AB S is midpoint of AB To find the sma ( γ) of AB see blow; In ΔACD, ∠D = 900 therefore; h2 = b2― (c/2+ γ)2 Pyth.theo. h2 = b2― c2/4 ― cγ ― γ2 e.q 1 In ΔBCD, ∠D = 900, therefore; h2 =a2― (c/2― γ)2 Pyth.theo. h2 =a2―c2/4 +cγ ― γ2 e.q.2 γ = (b2― a2)/2c combining e.q 1 and e.q.2 Similarly, smas of AC & BC are given below. Sma of BC is a/2 (i.e. α = a/2) (AC is ⊥ to BC & AM is the median of BC) Sma of AC is b/2 (i.e. β = b/2) (BC is ⊥ to AC & BN is the median of AC) The smas of side a and b are a/2 and b/2 respectively because the two legs themselves are altitudes of the triangle.Therefore,in this case, we should calculate the sma of the hypotenuse only.
So to prove the relation ― aα +cγ = bβ see blow; 2aα +2cγ +2bβ = 2a (a/2) + 2c (b2 – a2)/2c +2b (b/2) 2aα +2cγ + 2bβ = a2+ b2 – a2+ b2, where c2 = a2+b2 (Pyth.theo.) 2aα +2cγ +2bβ = 2c2 – 2a2, where c2 – a2 = 2bβ (from the above, β = ( c2 – a2 )/2b),therefore; 2aα +2cγ +2bβ = 4 bβ aα +cγ = bβ, hence it is proven.
Discussion
Many mathematical theorems on triangles discuss relationships among sides, angles or sides and angles of triangles. However, many of these theorems are on specific triangles only (e.g. Pythagoras theorem, double angle triangle theorem and triple angle triangle theorem). In contrast, the theorem this paper discovers is applicable to any triangle and this can be taken as a great advantage in many areas it is applicable to. Moreover, having known only three out of the six in the theorem, the remaining can be calculated. This is performed to make ‘’Solving the triangle’’ much easier.Although the theorems  sine and cosine rules are applicable to any triangle, still there is a difficulty to remember the sine or cosine of each angle to use such theorems on the triangle of interest. In contrast, Sma’s theorem simplifies this problem to some extent since it does not involve sine or cosine of any angle of the triangle of interest. As the “sma” of any side of the triangle, itself is a segment of that side, it can be more convenient to relate smas and sides than using other theorems on some special cases(e.g. if the hypotenuse and the sma of the hypotenuse are known then, the remaining legs of the Right angled triangle can be easily calculated). Therefore, here, it is suggested that the relation this article briefs can be applied for solving many important problems.
Conclusion
This theorem is applicable to any triangle and can simplify solving problems it is applicable to. This is a little challenging but this theorem is acceptable. This proven theorem is suggestive only not conclusive. Sam’s theorem shall be applied in the scientific field with experimental studies.
References [1] Bostock, L. and Chandler, S. Core Mathematics. Nelson Thornes: London, 2000, P.71 [2] Das, G.et al Topics in Mathematics, Kalyani Publisher: New Delhi, 2000, P.243
