Six New Theorems about Inscribed Triangle in a Circle

1 Six New Theorems about Inscribed Triangle in a Circle Salman Mahmud student of Government Azizul Haque College, Bogura, Bangladesh Abstract: In this paper we have proven 6 new theorems about inscribed triangle in a circle. If we extend the three medians of any inscribed triangle in a circle, the extended medians will intersect the circumference of the circle at three points. Connecting those three points we will get a new triangle. The main purpose of this paper is to describe the properties of the second triangle obtained by extending the medians of the first triangle. Keywords: New theorems, inscribed triangle in a circle, increasing medians. 1. Introduction There are a lot of theorems about inscribed triangle in a circle. In this study we have demonstrated 6 new theorems about inscribed triangle in a circle. If we extend the three medians of any inscribed triangle in a circle, the extended medians will intersect the circumference of the circle at three points. Connecting those three points we will get a new triangle. The main purpose of this paper is to describe the properties of the second triangle obtained by extending the medians of the first triangle. Here we have shown that if the first triangle is a scalene triangle then the second triangle will be a scalene triangle or isosceles triangle, if the first triangle is an isosceles triangle then the second triangle will also be an isosceles triangle and no matter the first isosceles triangle is acute or right or obtuse the second isosceles triangle will be always acute, if the first triangle is equilateral then the second triangle will also be equilateral, if the first triangle is an acute triangle then the second triangle will also be an acute triangle, if the first triangle is a right triangle then the second triangle will be an acute triangle, if the first triangle is an obtuse triangle then the second triangle will be acute or right or obtuse. 2. Main results In this paper we have mainly proved 6 theorems about inscribed triangle in a circle and the theorems are as follows: Theorem 1.1: If the medians of a scalene triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will be a scalene triangle or isosceles triangle. Theorem 1.2: If the medians of an isosceles triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will also be an isosceles triangle and no matter the first isosceles triangle is acute or right or obtuse the second triangle will always be an acute isosceles triangle. 2 Theorem 1.3: If the medians of an equilateral triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will also be an equilateral triangle. Theorem 1.4: If the medians of an acute triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will also be an acute triangle. Theorem 1.5: If the medians of a right triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will be an acute triangle. Theorem 1.6: If the medians of an obtuse triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will be acute or right or obtuse. 3. Proof of the theorems Here we will prove all the theorems one by one. Before we prove the theorems, it is necessary to generate the formulas to calculate the length of the sides of the second triangle obtained by extending the medians of the first inscribed triangle in a circle. In other words, If we extend the three medians of any inscribed triangle in a circle, the extended medians will intersect the circumference of the circle at three points. Connecting those three points of intersection we will get a new triangle. Now we have to generate the formulas to calculate the length of the sides of this triangle. In this case we will use a method. We will consider any side of the first triangle as the main side and the other two sides as the secondary side and secondary side . Then we will represent the lengths of secondary side and secondary side by using the length of main side. Suppose, ABC is a triangle. We consider AB as the main side, BC as the secondary side and AC as the secondary side . Now we can write BC = AB𝛼 where 𝛼 = BC/AB. Again, AC = AB𝛽 where 𝛽 = AC/AB. Notation 1.1: The Greek letter "𝜑” is used as the length of main side of any triangle. Definition 1.1: We call, 𝛼 is the ratio of secondary side and the main side and 𝛽is the ratio of secondary side and the main side. To prove our theorems we need to use some important theorems and the theorems are as follows: Theorem 2.1 (Apollonius’s theorem): The sum of the squares of any two sides of any triangle equals to twice its square on half of the third side, along with the twice of its square on the median bisecting the third side. Theorem 2.2 (Pythagorean theorem): In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides. Theorem 2.3: Two angles at the circumference subtended by the same arc are equal. Theorem 2.4: The centroid of a triangle divides each median in the ratio 2:1. 3 Theorem 2.5: The sum of any two sides of a triangle is greater than the third side. Figure 1: ABC is the first triangle inscribed in the circle AFBDCE. After extending the medians of the triangle ABC, they intersect at three points D,E and F of the circumference of the circle. Connecting D,E and E,F and F,D, the second triangle DEF has been obtained. Suppose, the main side of the triangle ABC is AB. The secondary side = BC = AB𝛼 and the secondary side = AC = AB𝛽. By using the Apollonius’s theorem we can write from the figure 1, 𝐴𝐵 + 𝐴𝐶 = 2(𝐴𝐻 + 𝐵𝐻 ) 𝐴𝐵 + 𝐴𝐶 ⇒ 𝐴𝐻 = − 𝐵𝐻 2 𝐴𝐵 + 𝐴𝐵 𝛽 𝐴𝐵 𝛼 ⇒ 𝐴𝐻 = − 2 4 𝐴𝐵 (2 + 2𝛽 − 𝛼 ) ⇒ 𝐴𝐻 = 4 𝐴𝐵 2 + 2𝛽 − 𝛼 𝐴𝐻 = 2 According to theorem 2.4 we can write, 𝐴𝐺: 𝐺𝐻 = 2: 1 ⇒ 𝐴𝐻: 𝐺𝐻 = 3: 1 𝐴𝐵 2 + 2𝛽 − 𝛼 1 × 3 2 𝐴𝐵 2 + 2𝛽 − 𝛼 𝐺𝐻 = 6 ⇒ 𝐺𝐻 = Look at the figure 1, we can see the two angles ∠𝐷𝐵𝐶 and ∠𝐷𝐴𝐶 are subtended by the same arc CD. According to theorem 2.3, ∠𝐷𝐵𝐶 = ∠𝐷𝐴𝐶 Again, the two angles ∠𝐴𝐷𝐵 and ∠𝐴𝐶𝐵 are subtended by the same arc AB. That’s why ∠𝐴𝐷𝐵 = ∠𝐴𝐶𝐵 4 Now in case of triangle BHD and triangle AHC, ∠𝐷𝐵𝐻 = ∠𝐻𝐴𝐶 and ∠𝐻𝐷𝐵 = ∠𝐴𝐶𝐻. So, ∆BHD and ∆ AHC are similar. That’s why, 𝐷𝐻 𝐵𝐻 = 𝐻𝐶 𝐴𝐻 ⇒ 𝐷𝐻 = 𝐷𝐻 = Now, × 𝐴𝐵𝛼 2 2 + 2𝛽 − 𝛼 𝐺𝐷 = 𝐺𝐻 + 𝐷𝐻 = 𝐴𝐵 2 + 2𝛽 − 𝛼 𝐴𝐵𝛼 + 6 2 2 + 2𝛽 − 𝛼 = 2𝐴𝐵 + 2𝐴𝐵𝛽 + 2𝐴𝐵𝛼 = 6 2 + 2𝛽 − 𝛼 𝐴𝐵(1 + 𝛽 + 𝛼 ) 3 2 + 2𝛽 − 𝛼 Again, according to the Apollonius’s theorem we can write from the figure 1, 𝐴𝐶 + 𝐵𝐶 = 2(𝐶𝐽 + 𝐴𝐽 ) 𝐴𝐶 + 𝐵𝐶 ⇒ 𝐶𝐽 = − 𝐴𝐽 2 𝐴𝐵 𝐴𝐵 𝛽 + 𝐴𝐵 𝛼 − ⇒ 𝐶𝐽 = 4 2 2𝐴𝐵 𝛽 + 2𝐴𝐵 𝛼 − 𝐴𝐵 ⇒ 𝐶𝐽 = 4 𝐴𝐵 2𝛽 + 2𝛼 − 1 𝐶𝐽 = 2 According to theorem 2.4, 𝐶𝐺: 𝐺𝐽 = 2: 1 ⇒ 𝐶𝐺: 𝐶𝐽 = 2: 3 𝐴𝐵 2𝛽 + 2𝛼 − 1 2 ⇒ 𝐶𝐺 = × 3 2 𝐴𝐵 2𝛽 + 2𝛼 − 1 𝐶𝐺 = 3 Again, we can see the two angles ∠𝐷𝐹𝐶 and ∠𝐷𝐴𝐶 are subtended by the same arc CD. According to theorem 2.3, ∠𝐷𝐹𝐶 = ∠𝐷𝐴𝐶. Similarly, the two angles ∠𝐴𝐶𝐹 and ∠𝐴𝐷𝐹 are subtended by the same arc AF. That’s why ∠𝐴𝐶𝐹 = ∠𝐴𝐷𝐹. Now is case of ∆FGD and ∆AGC, ∠𝐷𝐹𝐺 = ∠𝐺𝐴𝐶 and ∠𝐹𝐷𝐺 = ∠𝐺𝐶𝐴. So, ∆FGD and ∆AGC are similar. That’s why, 𝐷𝐹 𝐺𝐷 = 𝐴𝐶 𝐶𝐺 5 ⇒ 𝐷𝐹 = ⇒ 𝐷𝐹 = 𝐴𝐵(1 + 𝛽 + 𝛼 ) 3 2 + 2𝛽 − 𝛼 𝐷𝐹 = 𝐺𝐷 × 𝐴𝐶 𝐶𝐺 × 𝐴𝐵𝛽 × 3 𝐴𝐵 2𝛽 + 2𝛼 − 1 𝐴𝐵𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) Again, according to the Apollonius’s theorem we can write from the figure 1, 𝐴𝐵 + 𝐵𝐶 = 2(𝐵𝐼 𝐴𝐵 + 𝐵𝐶 ⇒ 𝐵𝐼 = 2 𝐴𝐵 + 𝐴𝐵 𝛼 ⇒ 𝐵𝐼 = 2 𝐴𝐵 2 + 2𝛼 𝐵𝐼 = 2 + 𝐴𝐼 ) − 𝐴𝐼 𝐴𝐵 𝛽 4 −𝛽 − Now, by using theorem 2.4 we can write, 𝐴𝐵 2 + 2𝛼 − 𝛽 2 × 3 2 𝐴𝐵 2 + 2𝛼 − 𝛽 = 3 𝐵𝐺 = In case of ∆AGB and ∆DGE, ∠𝐵𝐴𝐺 = ∠𝐷𝐸𝐺 and ∠𝐴𝐵𝐺 = ∠𝐸𝐷𝐺 (Theorem 2.3). So, ∆AGB and ∆DGE are similar. That’s why, 𝐷𝐸 𝐺𝐷 = 𝐴𝐵 𝐵𝐺 3 𝐴𝐵(1 + 𝛽 + 𝛼 ) × 𝐴𝐵 × ⇒ 𝐷𝐸 = 𝐴𝐵 2 + 2𝛼 − 𝛽 3 2 + 2𝛽 − 𝛼 𝐴𝐵(1 + 𝛽 + 𝛼 ) 𝐷𝐸 = (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) According to theorem 2.4, 𝐽𝐺: 𝐶𝐺 = 1: 2 𝐶𝐺 ⇒ 𝐽𝐺 = 2 𝐴𝐵 2𝛽 + 2𝛼 − 1 𝐽𝐺 = 6 Now, in case of ∆AFJ and ∆BCJ,∠𝐹𝐴𝐽 = ∠𝐵𝐶𝐽 and ∠𝐴𝐹𝐽 = ∠𝐽𝐵𝐶 (Theorem 2.3). So, ∆AFJ and ∆BCJ are similar. That’s why, 𝐹𝐽 𝐴𝐽 = 𝐵𝐽 𝐶𝐽 𝐴𝐽 × 𝐵𝐽 ⇒ 𝐹𝐽 = 𝐶𝐽 6 ⇒ 𝐹𝐽 = Now, = 𝐴𝐵 𝐴𝐵 2 × × 2 2 𝐴𝐵 2𝛽 + 2𝛼 − 1 𝐴𝐵 𝐹𝐽 = 2 2𝛽 + 2𝛼 − 1 𝐴𝐵 𝐹𝐺 = 𝐹𝐽 + 𝐽𝐺 2 2𝛽 + 2𝛼 − 1 = + 𝐴𝐵 2𝛽 + 2𝛼 − 1 6 𝐴𝐵(1 + 𝛽 + 𝛼 ) 3 2𝛽 + 2𝛼 − 1 In case of ∆EFG and ∆BCG, ∠𝐸𝐹𝐺 = ∠𝐶𝐵𝐺 and ∠𝐹𝐸𝐺 = ∠𝐵𝐶𝐺 (Theorem 2.3). So, ∆EFG and ∆BCG are similar. That’s why, 𝐸𝐹 𝐹𝐺 = 𝐵𝐶 𝐵𝐺 𝐹𝐺 × 𝐵𝐶 ⇒ 𝐸𝐹 = 𝐵𝐺 3 𝐴𝐵(1 + 𝛽 + 𝛼 ) × 𝐴𝐵𝛼 × ⇒ 𝐸𝐹 = 𝐴𝐵 2 + 2𝛼 − 𝛽 3 2𝛽 + 2𝛼 − 1 𝐴𝐵𝛼(1 + 𝛽 + 𝛼 ) 𝐸𝐹 = (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) So, we have calculated the length of all sides of the triangle DEF by using the triangle ABC. In this case we took AB as the main side. But we can take any side as the main side. Again, we have to apply these formulas for any triangle. So, we have to generate the generalized form of these formulas. Here we can see the side DE of the triangle DEF is on the opposite side of AB and we have considered AB as the main side of the triangle ABC. So, we can write that the formula to calculate the length of one side of the second triangle which is on the opposite side of the main side of the first triangle is 𝜑(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) Here, we used 𝜑 as the length of main side of any triangle. Again, The side EF of the triangle DEF is on the opposite side of BC and we have considered BC as the secondary side of the triangle ABC. So, we can write that the formula to calculate the length of one side of the second triangle which is on the opposite side of the secondary side of the first triangle is 𝜑𝛼(1 + 𝛽 + 𝛼 ) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) Again, The side DF of the triangle DEF is on the opposite side of AC and we have considered AC as the secondary side of the triangle ABC. So, we can write that the formula to calculate the length of one side of the second triangle which is on the opposite side of the secondary side of the first triangle is 7 𝜑𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) Theorem 1.1: If the medians of a scalene triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will be a scalene triangle or isosceles triangle. Proof: Figure 2: 𝐴 𝐵 𝐶 is a scalene triangle inscribed in the circle. After extending the medians of the triangle 𝐴 𝐵 𝐶 , they intersect at three points of the circumference of the circle. Connecting those three points of intersection, the triangle 𝐴 𝐵 𝐶 has been obtained. The triangle 𝐴 𝐵 𝐶 is a scalene triangle. Suppose, 𝐴 𝐵 < 𝐵 𝐶 < 𝐴 𝐶 , the main side = 𝐴 𝐵 , secondary side = 𝐵 𝐶 = 𝐴 𝐵 𝛼 and secondary side = 𝐴 𝐶 = 𝐴 𝐵 𝛽 . Here, 𝛼 = 𝐵 𝐶 ⁄𝐴 𝐵 and 𝐴 𝐵 < 𝐵 𝐶 , that’s why, 𝛼 > 1. Again, 𝛽 = 𝐴 𝐶 ⁄𝐴 𝐵 and 𝐴 𝐵 < 𝐴 𝐶 , that’s why, 𝛽 is also greater than 1. Again, 𝐵 𝐶 < 𝐴 𝐶 , that’s why, 𝛼 < 𝛽 . So, we can write, 1 < 𝛼 < 𝛽 . Now, by using the theorem 2.5 we can write, 𝐴 𝐵 + 𝐵 𝐶 > 𝐴 𝐶 ⇒ 𝐴 𝐵 +𝐴 𝐵 𝛼 >𝐴 𝐵 𝛽 ⇒1+𝛼 >𝛽 ⇒𝛼 >𝛽−1 And we know 𝛼 < 𝛽. So, we can write, 𝛽 − 1 < 𝛼 < 𝛽. Again, we have got, 1 < 𝛼 < 𝛽 Now, we can see, when 𝛽 > 1 and 𝛽 ≤ 2 , that time 𝛽 − 1 ≤ 1 and when 𝛽 > 2, that time 𝛽 − 1 > 1. So, when 1 < 𝛽 ≤ 2 , we can write, 𝛽−1≤1<𝛼 <𝛽 ⇒1<𝛼<𝛽 Again, when 𝛽 > 2 , we can write, 1< 𝛽−1<𝛼 <𝛽 ⇒ 𝛽−1<𝛼 <𝛽 Again, 1 + 𝛼 > 𝛽 and 𝛼 < 𝛽. So, we can write, 𝛼 < 𝛽 < 1 + 𝛼. From this inequality we can write, 𝛽 = 𝛼 + 𝑥 where 0 < 𝑥 < 1. Now, By using theorem 2.5, we can also write, 𝐴 𝐵 + 𝐴 𝐶 > 𝐵 𝐶 8 ⇒ 𝐴 𝐵 + 𝐴 𝐵 𝛽 > 𝐴 𝐵 𝛼 ⇒ 1 + 𝛽 > 𝛼. This inequality is not helpful. Because, 𝛼 is less than 𝛽. That’s why 𝛼 is obviously less than 1 + 𝛽. Again, 𝐴 𝐵 𝛽 + 𝐴 𝐵 𝛼 > 𝐴 𝐵 ⇒ 𝛼 > 1 − 𝛽. This inequality is also not much helpful. As 𝛽 > 1, that’s why 1 − 𝛽 is negative. But 𝛼 > 1. That’s why 𝛼 is greater than any negative number. Now, we have to prove that, the triangle 𝐴 𝐵 𝐶 is a scalene triangle or isosceles triangle. Suppose, 𝐴 𝐵 𝐶 is not a scalene triangle or isosceles triangle. Then it must be an equilateral triangle. Then, 𝐴 𝐵 = 𝐵 𝐶 and 𝐵 𝐶 = 𝐴 𝐶 and 𝐴 𝐵 = 𝐴 𝐶 . Here, by using the above formulas we can write, 𝐴 𝐵 = 𝐵 𝐶 = 𝐴 𝐶 = If, ⇒ 𝐴 𝐵 (1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) 𝐴 𝐵 𝛼(1 + 𝛽 + 𝛼 ) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) 𝐴 𝐵 𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) 𝐴 𝐵 (1 + 𝛽 + 𝛼 ) 𝐴 𝐵 =𝐵 𝐶 = (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) ⇒ 1 (2 + 2𝛽 − 𝛼 ) = 𝐴 𝐵 𝛼(1 + 𝛽 + 𝛼 ) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) 𝛼 (2𝛽 + 2𝛼 − 1) ⇒ 2𝛼 + 2𝛼 𝛽 − 𝛼 = 2𝛽 + 2𝛼 − 1 ⇒ 𝛽 (2𝛼 − 2) = 𝛼 − 1 ⇒𝛽= 𝛼 −1 2𝛼 − 2 Now, we have calculated that 𝛼 < 𝛽 < 1 + 𝛼. So, 𝛽 will be equal to if and only if Now, 𝛼< 𝛼 −1 < 1+𝛼 2𝛼 − 2 𝛼< 𝛼 −1 2𝛼 − 2 ⇒ 2𝛼 − 2𝛼 < 𝛼 − 1 ⇒ 𝛼 − 2𝛼 + 1 < 0 Here we can see there is no solution of this inequality. That’s why, 𝛼 − 2𝛼 + 1 ≮ 0. So, 9 That’s why, 𝛼≮ 𝛼 −1 2𝛼 − 2 𝛽≠ 𝛼 −1 2𝛼 − 2 So, 𝐴 𝐵 ≠ 𝐵 𝐶 . Thus triangle 𝐴 𝐵 𝐶 cannot be an equilateral triangle. It can be an isosceles triangle or scalene triangle. Now suppose, 𝐵 𝐶 =𝐴 𝐶 ⇒ 𝐴 𝐵 𝛼(1 + 𝛽 + 𝛼 ) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) ⇒ 𝛼 (2 + 2𝛼 − 𝛽 ) = = 𝐴 𝐵 𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) 𝛽 (2 + 2𝛽 − 𝛼 ) ⇒ 2𝛼 + 2𝛼 𝛽 − 𝛼 = 2𝛽 + 2𝛼 𝛽 − 𝛽 ⇒ 2𝛼 − 𝛼 − 2𝛽 + 𝛽 = 0 Now, if we use 𝛽 = 𝛼 + 𝑥 where 0 < 𝑥 < 1, then, 2𝛼 − 𝛼 − 2(𝛼 + 𝑥) + (𝛼 + 𝑥) 2𝛼 − 𝛼 − 2𝛼 − 4𝛼𝑥 − 2𝑥 + 𝛼 + 4𝛼 𝑥 + 6𝛼 𝑥 + 4𝛼𝑥 + 𝑥 = 0 ⇒ 4𝛼 𝑥 − 4𝛼𝑥 + 6𝛼 𝑥 − 2𝑥 + 4𝛼𝑥 + 𝑥 = 0 As 𝛼 > 1 and 0 < 𝑥 < 1, that’s why, 4𝛼 𝑥 > 4𝛼𝑥 and 6𝛼 𝑥 > 2𝑥 . So, 4𝛼 𝑥 − 4𝛼𝑥 > 0 and 6𝛼 𝑥 − 2𝑥 > 0. That’s why, 4𝛼 𝑥 − 4𝛼𝑥 + 6𝛼 𝑥 − 2𝑥 + 4𝛼𝑥 + 𝑥 > 0. So, we can see, 4𝛼 𝑥 − 4𝛼𝑥 + 6𝛼 𝑥 − 2𝑥 + 4𝛼𝑥 + 𝑥 ≠ 0 So, 𝐵 𝐶 ≠ 𝐴 𝐶 . Thus, triangle 𝐴 𝐵 𝐶 will be an isosceles triangle if and only if 𝐴 𝐵 = 𝐴 𝐶 and the triangle 𝐴 𝐵 𝐶 will be a scalene triangle if 𝐴 𝐵 ≠ 𝐴 𝐶 . Now suppose, ⇒ 𝐴 𝐵 (1 + 𝛽 + 𝛼 ) 𝐴 𝐵 =𝐴 𝐶 (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) ⇒ 1 (2 + 2𝛼 − 𝛽 ) = = 𝐴 𝐵 𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) 𝛽 (2𝛽 + 2𝛼 − 1) ⇒ 2𝛽 + 2𝛼 − 1 = 2𝛽 + 2𝛼 𝛽 − 𝛽 ⇒ 𝛼 (2𝛽 − 2) = 𝛽 − 1 ⇒𝛼= 𝛽 −1 2𝛽 − 2 10 Now, we have calculated that, 1 < 𝛼 < 𝛽 when 1 < 𝛽 ≤ 2 and 𝛽 − 1 < 𝛼 < 𝛽 when 𝛽 > 2. So, 𝛼 will be equal to 1< Or, if, 𝛽 −1 <𝛽 2𝛽 − 2 𝛽 −1 < 𝛽 when 𝛽 > 2 2𝛽 − 2 𝛽−1< Suppose, 1< Now, when 1 < 𝛽 ≤ 2 𝛽 −1 <𝛽 2𝛽 − 2 1< when 1 < 𝛽 ≤ 2 𝛽 −1 2𝛽 − 2 ⇒ 2𝛽 − 2 < 𝛽 − 1 ⇒ 𝛽 − 2𝛽 + 1 > 0 Here we can see the solution of this inequality is, 𝛽 = (−∞, −1) ∪ (−1,1) ∪ (1, ∞). But in this case 𝛽 > 1 and 𝛽 ≤ 2. So, the acceptable solution of this inequality is, 𝛽 = {(−∞, −1) ∪ (−1,1) ∪ (1, ∞)} ∩ (1,2] = (1,2] So,1 < Again, is true when 1 < 𝛽 ≤ 2. 𝛽 −1 <𝛽 2𝛽 − 2 ⇒ 𝛽 − 1 < 2𝛽 − 2𝛽 ⇒ 𝛽 − 2𝛽 + 1 > 0 Now, we get the same inequality. So, why, 1< < 𝛽, this is true when 1 < 𝛽 ≤ 2. That’s 𝛽 −1 < 𝛽 , This is true when 1 < 𝛽 ≤ 2 2𝛽 − 2 Now, if we take 𝛼 = when 1 < 𝛽 ≤ 2 , then 𝐴 𝐵 = 𝐴 𝐶 . So, the triangle 𝐴 𝐵 𝐶 will be an isosceles triangle. But if we take, 1<𝛼< 𝛽 −1 2𝛽 − 2 or, 𝛽 −1 <𝛼<𝛽 2𝛽 − 2 when 1 < 𝛽 ≤ 2 Then 𝐴 𝐵 ≠ 𝐴 𝐶 . So, the triangle 𝐴 𝐵 𝐶 will be a scalene triangle. 11 Again suppose, 𝛽−1< Now, 𝛽 −1 < 𝛽 when 𝛽 > 2 2𝛽 − 2 𝛽−1< 𝛽 −1 2𝛽 − 2 ⇒ 2𝛽 − 4𝛽 + 2𝛽 − 2𝛽 + 4𝛽 − 2 < 𝛽 − 1 ⇒ 𝛽 − 4𝛽 + 4𝛽 − 1 < 0 Here we can see the solution of this inequality is, 𝛽 = −1,2 − √3 ∪ 1,2 + √3 . But in this case 𝛽 > 2. So, the acceptable solution of this inequality is, 𝛽 = −1,2 − √3 ∪ 1,2 + √3 ∩ (2, ∞) = 2,2 + √3 Again, 𝛽 −1 <𝛽 2𝛽 − 2 ⇒ 𝛽 − 1 < 2𝛽 − 2𝛽 ⇒ 𝛽 − 2𝛽 + 1 > 0 Here we can see the solution of this inequality is, 𝛽 = (−∞, −1) ∪ (−1,1) ∪ (1, ∞). But in this case 𝛽 > 2. So, the acceptable solution of this inequality is, 𝛽 = {(−∞, −1) ∪ (−1,1) ∪ (1, ∞)} ∩ (2, ∞) = (2, ∞) Now, < 𝛽 , this will be true when 𝛽 = 2,2 + √3 ∩ (2, ∞)= 2,2 + √3 𝛽−1< Now, if we take 𝛼 = when 2 < 𝛽 < 2 + √3, then 𝐴 𝐵 = 𝐴 𝐶 . So, the triangle 𝐴 𝐵 𝐶 will be an isosceles triangle. But if we take, 𝛽−1<𝛼 < 𝛽 −1 2𝛽 − 2 or, 𝛽 −1 <𝛼<𝛽 2𝛽 − 2 when 2 < 𝛽 < 2 + √3 Then 𝐴 𝐵 ≠ 𝐴 𝐶 . So, the triangle 𝐴 𝐵 𝐶 will be a scalene triangle. Again, when 𝛽 ≥ 2 + √3 then, So, 𝛼 ≠ 𝛽−1≮ 𝛽 −1 2𝛽 − 2 . That’s why, 𝐴 𝐵 ≠ 𝐴 𝐶 . So, the triangle 𝐴 𝐵 𝐶 will be a scalene triangle when 𝛽 ≥ 2 + √3 . Thus, it is proved that the triangle 𝐴 𝐵 𝐶 can be an isosceles triangle or scalene triangle. Theorem 1.2: If the medians of an isosceles triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will also be an isosceles triangle and no matter the first isosceles triangle is acute or right or obtuse the second triangle will always be an acute isosceles triangle. 12 Figure 3: 𝐴 𝐵 𝐶 is an isosceles triangle inscribed in the circle. After extending the medians of the triangle 𝐴 𝐵 𝐶 , they intersect at three points of the circumference of the circle. Connecting those three points of intersection, the triangle 𝐴 𝐵 𝐶 has been obtained. The triangle 𝐴 𝐵 𝐶 is an isosceles triangle. Suppose, 𝐴 𝐵 = 𝐴 𝐶 ≠ 𝐵 𝐶 , the main side = 𝐵 𝐶 , secondary side = 𝐴 𝐵 = 𝐵 𝐶 𝛼 and secondary side = 𝐴 𝐶 = 𝐵 𝐶 𝛽 . Here, 𝛼 = 𝐴 𝐵 /𝐵 𝐶 and 𝛽 = 𝐴 𝐶 /𝐵 𝐶 . Again, 𝐴 𝐵 = 𝐴 𝐶 . That’s why, 𝛼 = 𝛽 . Now, by using the theorem 2.5 we can write, 𝐴 𝐵 +𝐴 𝐶 >𝐵 𝐶 ⇒ 2𝐵 𝐶 𝛼 > 𝐵 𝐶 [𝛽 = 𝛼] 1 𝛼> 2 Now, by using the formulas to calculate the length of the sides of the second triangle we can write, 𝐵 𝐶 𝛼(1 + 𝛽 + 𝛼 ) 𝐴 𝐵 = (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) = 𝐵 𝐶 𝛼(1 + 𝛼 + 𝛼 ) (2𝛼 + 2𝛼 − 1)(2 + 2𝛼 − 𝛼 ) = Again, 𝐵 𝐶 = Again, = 𝐴 𝐶 = = [𝛽 = 𝛼] 𝐵 𝐶 𝛼(1 + 2𝛼 ) (4𝛼 − 1)(2 + 𝛼 ) 𝐵 𝐶 (1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) 𝐵 𝐶 (1 + 2𝛼 ) (2 + 𝛼 )(2 + 𝛼 ) [𝛽 = 𝛼] 𝐵 𝐶 𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) 𝐵 𝐶 𝛼(1 + 2𝛼 ) (4𝛼 − 1)(2 + 𝛼 ) [𝛽 = 𝛼] Now we can see, 𝐴 𝐵 = 𝐴 𝐶 . So, The triangle 𝐴 𝐵 𝐶 is an isosceles triangle. Now, we have to prove that triangle 𝐴 𝐵 𝐶 is an acute triangle. If the triangle 𝐴 𝐵 𝐶 is an acute triangle, then all the angles of this triangle will be acute angle. Again, if the triangle 𝐴 𝐵 𝐶 is an acute triangle, then we can write, 13 𝐴 𝐵 + 𝐵 𝐶 > 𝐴 𝐶 and 𝐴 𝐵 + 𝐴 𝐶 > 𝐵 𝐶 and 𝐵 𝐶 + 𝐴 𝐶 > 𝐴 𝐵 . Now suppose, triangle 𝐴 𝐵 𝐶 is not an acute triangle. Then it can be a right triangle or obtuse triangle. So, if the triangle 𝐴 𝐵 𝐶 is a right triangle or obtuse triangle, then we can write, or, 𝐴 𝐵 + 𝐴 𝐶 ≤ 𝐵 𝐶 . 𝐴 𝐵 + 𝐵 𝐶 ≤ 𝐴 𝐶 or, 𝐵 𝐶 + 𝐴 𝐶 ≤ 𝐴 𝐵 Now suppose, 𝐴 𝐵 +𝐵 𝐶 ≤𝐴 𝐶 ⇒𝐴 𝐵 +𝐵 𝐶 −𝐴 𝐶 ≤0 ⇒ 𝐵 𝐶 ≤0 But this is not true because 𝐵 𝐶 cannot be negative or 0. That’s why, 𝐵 𝐶 ≰ 0. So, 𝐵 𝐶 > 0 and 𝐴 𝐵 + 𝐵 𝐶 > 𝐴 𝐶 . Similarly, 𝐵 𝐶 + 𝐴 𝐶 ≰ 𝐴 𝐵 . That’s why, 𝐵 𝐶 + 𝐴 𝐶 > 𝐴 𝐵 . Again suppose, ⇒ 𝐵 𝐶 𝛼(1 + 2𝛼 ) (4𝛼 − 1)(2 + 𝛼 ) ⇒ +𝐴 𝐶 𝐴 𝐵 + ≤𝐵 𝐶 𝐵 𝐶 𝛼(1 + 2𝛼 ) (4𝛼 − 1)(2 + 𝛼 ) ≤ 𝐵 𝐶 (1 + 2𝛼 ) (2 + 𝛼 )(2 + 𝛼 ) 𝐵 𝐶 (1 + 2𝛼 ) (4𝛼 + 2𝛼 − 4𝛼 + 1) ≤0 (4𝛼 − 1)(2 + 𝛼 ) ⇒ 𝐵 𝐶 (1 + 2𝛼 ) (2𝛼 + 1) ≤0 (4𝛼 − 1)(2 + 𝛼 ) Here, (2 + 𝛼 ) > 0 because 𝛼 > . That’s why, we will get the real value of (4𝛼 − 1)(2 + 𝛼 ) if and only if (4𝛼 − 1) > 0. So, (4𝛼 − 1) must be positive. So, in the denominator we have positive value. Again, we can see in the numerator we have 𝐵 𝐶 (1 + 2𝛼 ) (2𝛼 + 1). As 𝛼 > That’s why, ( )( ) , so 𝐵 𝐶 (1 + 2𝛼 ) (2𝛼 + 1) is positive. cannot be negative or 0. So, 𝐵 𝐶 (1 + 2𝛼 ) (2𝛼 + 1) ≰0 (4𝛼 − 1)(2 + 𝛼 ) That’s why, 𝐴 𝐵 + 𝐴 𝐶 ≰ 𝐵 𝐶 . So, 𝐴 𝐵 + 𝐴 𝐶 > 𝐵 𝐶 . Thus, triangle 𝐴 𝐵 𝐶 cannot be a right triangle or obtuse triangle. It must be an isosceles triangle. Theorem 1.3: If the medians of an equilateral triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will also be an equilateral triangle. 14 Figure 4: 𝐴 𝐵 𝐶 is an equilateral triangle inscribed in the circle. After extending the medians of the triangle 𝐴 𝐵 𝐶 , they intersect at three points of the circumference of the circle. Connecting those three points of intersection, the triangle 𝐴 𝐵 𝐶 has been obtained. The triangle 𝐴 𝐵 𝐶 is an equilateral triangle. So, 𝐴 𝐵 = 𝐵 𝐶 = 𝐴 𝐶 . Suppose, the main side = 𝐴 𝐵 , secondary side = 𝐵 𝐶 = 𝐴 𝐵 𝛼 and secondary side = 𝐴 𝐶 = 𝐴 𝐵 𝛽 . Here, 𝛼 = 𝐵 𝐶 ⁄𝐴 𝐵 = 1 because 𝐴 𝐵 = 𝐵 𝐶 . Again, 𝛽 = 𝐴 𝐶 ⁄𝐴 𝐵 = 1 because 𝐴 𝐵 = 𝐴 𝐶 . Now, by using the formulas to calculate the length of the sides of the second triangle we can write, 𝐴 𝐵 (1 + 𝛽 + 𝛼 ) 𝐴 𝐵 = (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) = 𝐴 𝐵 ×3 𝐵 𝐶 = = =𝐴 𝐵 [𝛼 = 𝛽 = 1] 𝐴 𝐵 𝛼(1 + 𝛽 + 𝛼 ) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) 𝐴 𝐵 ×3 𝐴 𝐶 = = √9 √9 =𝐴 𝐵 [𝛼 = 𝛽 = 1] 𝐴 𝐵 𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) 𝐴 𝐵 ×3 √9 =𝐴 𝐵 [𝛼 = 𝛽 = 1] Now we get, 𝐴 𝐵 = 𝐵 𝐶 = 𝐴 𝐶 = 𝐴 𝐵 . That’s why, triangle 𝐴 𝐵 𝐶 equilateral triangle. is an Now we have to prove an important thing to prove the last three theorems. Suppose, ABC is a scalene triangle inscribed in a circle. By extending the medians of the triangle ABC we get the second triangle DEF. Now we have to prove that if we consider the shortest side of the triangle ABC as the main side, medium side as the secondary side and longest side as the secondary side then the sum of the square of the side of triangle DEF which is on the opposite side of the main side of the triangle ABC and the side which is on the opposite side of the secondary side of the triangle ABC is greater than the square of the side of the triangle DEF which is on the opposite side of the secondary side of the triangle ABC. Again, we have to prove that the sum of the square of the side 15 of triangle DEF which is on the opposite side of the secondary side of the triangle ABC and the side which is on the opposite side of the secondary side of the triangle ABC is greater than the square of the side of the triangle DEF which is on the opposite side of the main side of the triangle ABC. Now suppose, in case of triangle ABC, 𝐴𝐵 < 𝐵𝐶 < 𝐴𝐶 , the main side = 𝐴𝐵, the secondary side = 𝐵𝐶 = 𝐴𝐵𝛼 and the secondary side = 𝐴𝐶 = 𝐴𝐵𝛽. So we can write (look at the proof of the theorem 1.1), (a) 1 < 𝛼 < 𝛽 (b) 𝛼 < 𝛽 < 1 + 𝛼. So we can write, 𝛽 = 𝛼 + 𝑥 where 0 < 𝑥 < 1. Now suppose, DE is on the opposite side of the main side of triangle ABC. So, 𝐴𝐵(1 + 𝛽 + 𝛼 ) 𝐷𝐸 = (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) Again suppose, EF is on the opposite side of the secondary side of triangle ABC. So, 𝐴𝐵𝛼(1 + 𝛽 + 𝛼 ) 𝐸𝐹 = (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) Again suppose, DF is on the opposite side of the secondary side of triangle ABC. So, 𝐴𝐵𝛽(1 + 𝛽 + 𝛼 ) 𝐷𝐹 = (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) Now, we have to prove that, (a) 𝐷𝐸 + 𝐸𝐹 > 𝐷𝐹 (b) 𝐸𝐹 + 𝐷𝐹 > 𝐷𝐸 . Suppose, 𝐷𝐸 + 𝐸𝐹 ≯ 𝐷𝐹 . So, 𝐷𝐸 + 𝐸𝐹 ≤ 𝐷𝐹 𝐴𝐵𝛼(1 + 𝛽 + 𝛼 ) 𝐴𝐵(1 + 𝛽 + 𝛼 ) + ⇒ (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) ≤ ⇒ 𝐴𝐵𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) 𝐴𝐵 (1 + 𝛽 + 𝛼 ) {2𝛽 + 2𝛼 − 1 + 𝛼 (2 + 2𝛽 − 𝛼 ) − 𝛽 (2 + 2𝛼 − 𝛽 )} ≤0 (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 )(2𝛽 + 2𝛼 − 1) As 1 < 𝛼 < 𝛽, that’s why (2 + 2𝛽 − 𝛼 ) > 0. So, we will get the real value (greater than 0) of (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) if and only if, (2 + 2𝛼 − 𝛽 ) > 0. Again, (2𝛽 + 2𝛼 − 1) > 0 because 1 < 𝛼 < 𝛽. So, (2 + 2𝛽 − 𝛼 ), (2 + 2𝛼 − 𝛽 ) and (2𝛽 + 2𝛼 − 1) are positive. Again, (1 ) 𝐴𝐵 +𝛽 +𝛼 is also positive. That’s why, 𝐷𝐸 + 𝐸𝐹 ≤ 𝐷𝐹 , this will be true if, 2𝛽 + 2𝛼 − 1 + 𝛼 (2 + 2𝛽 − 𝛼 ) − 𝛽 (2 + 2𝛼 − 𝛽 ) ≤ 0 ⇒ 4𝛼 − 1 + 𝛽 − 𝛼 ≤ 0 As 1 < 𝛼 < 𝛽, that’s why 4𝛼 − 1 > 0 and 𝛽 − 𝛼 > 0. So, 4𝛼 − 1 + 𝛽 − 𝛼 > 0. That’s why, 4𝛼 − 1 + 𝛽 − 𝛼 ≰ 0. So, 𝐷𝐸 + 𝐸𝐹 ≰ 𝐷𝐹 . So, it is proved that 𝐷𝐸 + 𝐸𝐹 > 𝐷𝐹 Again suppose, 𝐸𝐹 + 𝐷𝐹 ≯ 𝐷𝐸 . So, 𝐸𝐹 + 𝐷𝐹 ≤ 𝐷𝐸 𝐴𝐵𝛽(1 + 𝛽 + 𝛼 ) 𝐴𝐵𝛼(1 + 𝛽 + 𝛼 ) + ⇒ (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) ≤ 𝐴𝐵(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) 16 ⇒ 𝐴𝐵 (1 + 𝛽 + 𝛼 ) {𝛼 (2 + 2𝛽 − 𝛼 ) + 𝛽 (2 + 2𝛼 − 𝛽 ) − 2𝛽 − 2𝛼 + 1} ≤0 (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 )(2𝛽 + 2𝛼 − 1) As we have proved that (2 + 2𝛽 − 𝛼 ), (2 + 2𝛼 − 𝛽 ) , (2𝛽 + 2𝛼 − 1) and 𝐴𝐵 (1 + 𝛽 + 𝛼 ) are positive. That’s why, 𝐸𝐹 + 𝐷𝐹 ≤ 𝐷𝐸 , this will be true if , 𝛼 (2 + 2𝛽 − 𝛼 ) + 𝛽 (2 + 2𝛼 − 𝛽 ) − 2𝛽 − 2𝛼 + 1 ≤ 0 ⇒ 4𝛼 𝛽 + 1 − 𝛼 − 𝛽 ≤ 0 ⇒ 4𝛼 (𝛼 + 𝑥) + 1 − 𝛼 − (𝛼 + 𝑥) ≤ 0 [ 𝛽 = 𝛼 + 𝑥 where 0 < 𝑥 < 1] ⇒ 4𝛼 + 8𝛼 𝑥 + 4𝛼 𝑥 + 1 − 𝛼 − 𝛼 − 4𝛼 𝑥 − 6𝛼 𝑥 − 4𝛼𝑥 − 𝑥 ≤ 0 ⇒ 2𝛼 − 2𝛼 𝑥 + 4𝛼 𝑥 − 4𝛼𝑥 + 1 − 𝑥 ≤ 0 Here, 𝛼 > 1 and 0 < 𝑥 < 1, that’s why 4𝛼 𝑥 > 4𝛼𝑥 , 1 > 𝑥 and 2𝛼 > 2𝛼 𝑥 . So, 4𝛼 𝑥 − 4𝛼𝑥 > 0, 1 − 𝑥 > 0 and 2𝛼 − 2𝛼 𝑥 > 0. That’s why, 2𝛼 − 2𝛼 𝑥 + 4𝛼 𝑥 − 4𝛼𝑥 + 1 − 𝑥 > 0 So, 2𝛼 − 2𝛼 𝑥 + 4𝛼 𝑥 − 4𝛼𝑥 + 1 − 𝑥 ≰ 0 That’s why, 𝐸𝐹 + 𝐷𝐹 ≰ 𝐷𝐸 . So, it is proved that, 𝐸𝐹 + 𝐷𝐹 > 𝐷𝐸 Theorem 1.4: If the medians of an acute triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will also be an acute triangle. Figure 5: 𝐴 𝐵 𝐶 is an acute triangle inscribed in the circle. After extending the medians of the triangle 𝐴 𝐵 𝐶 , they intersect at three points of the circumference of the circle. Connecting those three points of intersection, the triangle 𝐴 𝐵 𝐶 has been obtained. As the triangle 𝐴 𝐵 𝐶 is an acute triangle, so it can be an equilateral or isosceles triangle or scalene triangle. If 𝐴 𝐵 𝐶 is an equilateral triangle, then the triangle 𝐴 𝐵 𝐶 will be 17 an equilateral triangle and an equilateral triangle is an acute triangle. Again, if 𝐴 𝐵 𝐶 is an isosceles triangle, then the triangle 𝐴 𝐵 𝐶 will be an acute isosceles triangle according to theorem 1.2. Now we have to prove that if 𝐴 𝐵 𝐶 is an scalene triangle, then the triangle 𝐴 𝐵 𝐶 will be an acute triangle. Suppose, in case of triangle 𝐴 𝐵 𝐶 , 𝐴 𝐵 < 𝐵 𝐶 < 𝐴 𝐶 , the main side = 𝐴 𝐵 , secondary side = 𝐵 𝐶 = 𝐴 𝐵 𝛼 and secondary side = 𝐴 𝐶 = 𝐴 𝐵 𝛽 . Here, 𝛼 = 𝐵 𝐶 ⁄𝐴 𝐵 and 𝐴 𝐵 < 𝐵 𝐶 , that’s why, 𝛼 > 1. Again, 𝛽 = 𝐴 𝐶 ⁄𝐴 𝐵 and 𝐴 𝐵 < 𝐴 𝐶 , that’s why, 𝛽 is also greater than 1. Again, 𝐵 𝐶 < 𝐴 𝐶 , that’s why, 𝛼 < 𝛽 . So, we can write, 1<𝛼<𝛽 ⇒1<𝛼 <𝛽 As the triangle 𝐴 𝐵 𝐶 is an acute triangle, that’s why all the angles of this triangle are acute angles. So, 𝐴 𝐵 +𝐵 𝐶 >𝐴 𝐶 ⇒𝐴 𝐵 +𝐴 𝐵 𝛼 >𝐴 𝐵 𝛽 ⇒𝛼 >𝛽 −1 𝛼 > 𝛽 − 2𝛽 + 1 So we can write, 𝛽 − 2𝛽 + 1 < 𝛼 < 𝛽 . Now, we can see, when 𝛽 > 1 and 𝛽 ≤ √2 , that time 𝛽 − 2𝛽 + 1 ≤ 1 and when 𝛽 > √2, that time 𝛽 − 2𝛽 + 1 > 1. So when 1< 𝛽 ≤ √2, we can write, 𝛽 − 2𝛽 + 1 ≤ 1 < 𝛼 < 𝛽 ⇒ 1<𝛼 <𝛽 Again, when 𝛽 > √2, we can write, 1 < 𝛽 − 2𝛽 + 1 < 𝛼 < 𝛽 ⇒ 𝛽 − 2𝛽 + 1 < 𝛼 < 𝛽 Now, by using the formulas to calculate the length of the sides of the second triangle we can write, 𝐴 𝐵 = 𝐵 𝐶 = 𝐴 𝐶 = 𝐴 𝐵 (1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) 𝐴 𝐵 𝛼(1 + 𝛽 + 𝛼 ) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) 𝐴 𝐵 𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) Here, we have considered the shortest side of the triangle 𝐴 𝐵 𝐶 as the main side, medium side as the secondary side and longest side as the secondary side . That’s why, 𝐴 𝐵 + 𝐵 𝐶 > 𝐴 𝐶 and 𝐵 𝐶 + 𝐴 𝐶 > 𝐴 𝐵 . So, the triangle 𝐴 𝐵 𝐶 will be an acute triangle if and only if, 𝐴 𝐵 + 𝐴 𝐶 > 𝐵 𝐶 . Now, suppose the triangle 𝐴 𝐵 𝐶 is not an acute triangle. So, it can be a right triangle or obtuse triangle. At first, suppose 𝐴 𝐵 𝐶 is a right triangle. The triangle 𝐴 𝐵 𝐶 will be a right triangle if and only if, 𝐴 𝐵 +𝐴 𝐶 =𝐵 𝐶 ⇒ 𝐴 𝐵 (1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) + 𝐴 𝐵 𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) 18 = ⇒ 𝐴 𝐵 𝛼(1 + 𝛽 + 𝛼 ) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) 𝐴𝐵 (1 + 𝛽 + 𝛼 ) {2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 )} =0 (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 )(2𝛽 + 2𝛼 − 1) As we have proved that (2 + 2𝛽 − 𝛼 ), (2 + 2𝛼 − 𝛽 ) , (2𝛽 + 2𝛼 − 1) and 𝐴𝐵 (1 + 𝛽 + 𝛼 ) are greater than 0 when 1< 𝛼 < 𝛽. That’s why, 𝐴 𝐵 + 𝐴 𝐶 = 𝐵 𝐶 , this will be true if, 2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 ) = 0 ⇒ 𝛼 = 𝛽 − 4𝛽 + 1 We have calculated that 1 < 𝛼 < 𝛽 when 1< 𝛽 ≤ √2 and 𝛽 − 2𝛽 + 1 < 𝛼 < 𝛽 when 𝛽 > √2. So, 𝛽 − 4𝛽 + 1 will be equal to 𝛼 if, Or, 1 < 𝛽 − 4𝛽 + 1 < 𝛽 when 1 < 𝛽 ≤ √2 𝛽 − 2𝛽 + 1 < 𝛽 − 4𝛽 + 1 < 𝛽 when 𝛽 > √2 Now suppose, 1 < 𝛽 − 4𝛽 + 1 < 𝛽 when 1 < 𝛽 ≤ √2. So, 1 < 𝛽 − 4𝛽 + 1 ⇒ 𝛽 − 4𝛽 > 0 Here we can see the solution of this inequality is, 𝛽 = (−∞, −2) ∪ (2, ∞) . But in this case 1 < 𝛽 ≤ √2. So, the acceptable solution of this inequality is, 𝛽 = {(−∞, −2) ∪ (2, ∞)} ∩ (1, √2] =∅ So, there is no acceptable solution of this inequality. That’s why, 1 ≮ 𝛽 − 4𝛽 + 1 So, 𝛼 ≠ 𝛽 − 4𝛽 + 1 when 1 < 𝛽 ≤ √2 . That’s why, triangle 𝐴 𝐵 𝐶 cannot be a right triangle when 1 < 𝛽 ≤ √2 . Again suppose, 𝛽 − 2𝛽 + 1 < 𝛽 − 4𝛽 + 1 < 𝛽 when 𝛽 > √2 So, 𝛽 − 2𝛽 + 1 < 𝛽 − 4𝛽 + 1 ⇒ −2𝛽 < −4𝛽 But in this case 𝛽 > √2, that’s why, −2𝛽 ≮ −4𝛽 . So, 𝛽 − 2𝛽 + 1 ≮ 𝛽 − 4𝛽 + 1. That’s why triangle 𝐴 𝐵 𝐶 cannot be a right triangle when 𝛽 > √2. Now suppose, the triangle 𝐴 𝐵 𝐶 is an obtuse triangle. The triangle 𝐴 𝐵 𝐶 will be an obtuse triangle if and only if, 𝐴 𝐵 +𝐴 𝐶 <𝐵 𝐶 ⇒ 𝐴𝐵 (1 + 𝛽 + 𝛼 ) {2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 )} <0 (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 )(2𝛽 + 2𝛼 − 1) 19 Here, 𝐴 𝐵 +𝐴 𝐶 < 𝐵 𝐶 , this will be true if, 2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 ) < 0 ⇒ 𝛼 < 𝛽 − 4𝛽 + 1 Now, if 𝛽 − 4𝛽 + 1 is greater than 𝛼 , then 𝛽 − 4𝛽 + 1 must be greater than 1 when 1 < 𝛽 ≤ √2 and 𝛽 − 4𝛽 + 1 must be greater than 𝛽 − 2𝛽 + 1 when 𝛽 > √2. But we have proved that, 1 ≮ 𝛽 − 4𝛽 + 1 when 1 < 𝛽 ≤ √2 And, 𝛽 − 2𝛽 + 1 ≮ 𝛽 − 4𝛽 + 1 when 𝛽 > √2 So, 𝐴 𝐵 + 𝐴 𝐶 ≮ 𝐵 𝐶 . That’s why triangle 𝐴 𝐵 𝐶 cannot be an obtuse triangle. As the triangle 𝐴 𝐵 𝐶 cannot be a right triangle or obtuse triangle, so it must be an acute triangle. Theorem 1.5: If the medians of a right triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will be an acute triangle. Figure 6: 𝐴 𝐵 𝐶 is a right triangle inscribed in the circle. After extending the medians of the triangle 𝐴 𝐵 𝐶 , they intersect at three points of the circumference of the circle. Connecting those three points of intersection, the triangle 𝐴 𝐵 𝐶 has been obtained. As the triangle 𝐴 𝐵 𝐶 is a right triangle, so it can be an isosceles triangle or scalene triangle. If 𝐴 𝐵 𝐶 is an isosceles triangle, then the triangle 𝐴 𝐵 𝐶 will be an acute isosceles triangle according to theorem 1.2. Now we have to prove that if 𝐴 𝐵 𝐶 is an scalene triangle, then the triangle 𝐴 𝐵 𝐶 will be an acute triangle. Suppose, in case of triangle 𝐴 𝐵 𝐶 , 𝐴 𝐵 < 𝐵 𝐶 < 𝐴 𝐶 , the main side = 𝐴 𝐵 , secondary side = 𝐵 𝐶 = 𝐴 𝐵 𝛼 and secondary side = 𝐴 𝐶 = 𝐴 𝐵 𝛽 . Here, 𝛼 = 𝐵 𝐶 ⁄𝐴 𝐵 and 𝐴 𝐵 < 𝐵 𝐶 , that’s why, 𝛼 > 1 . Again, 𝛽 = 𝐴 𝐶 ⁄𝐴 𝐵 and 𝐴 𝐵 < 𝐴 𝐶 , that’s why, 𝛽 is also greater than 1. Again, 𝐵 𝐶 < 𝐴 𝐶 , that’s why, 𝛼 < 𝛽 . So, we can write, 1<𝛼<𝛽 As the triangle 𝐴 𝐵 𝐶 is a right triangle, so 𝐴 𝐵 +𝐵 𝐶 =𝐴 𝐶 20 ⇒𝐴 𝐵 +𝐴 𝐵 𝛼 =𝐴 𝐵 𝛽 ⇒𝛼 =𝛽 −1 Now, by using the formulas to calculate the length of the sides of the second triangle we can write, 𝐴 𝐵 = 𝐵 𝐶 = 𝐴 𝐶 = 𝐴 𝐵 (1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) 𝐴 𝐵 𝛼(1 + 𝛽 + 𝛼 ) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) 𝐴 𝐵 𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) Here, we have considered the shortest side of the triangle 𝐴 𝐵 𝐶 as the main side, medium side as the secondary side and longest side as the secondary side . That’s why, 𝐴 𝐵 + 𝐵 𝐶 > 𝐴 𝐶 and 𝐵 𝐶 + 𝐴 𝐶 > 𝐴 𝐵 . So, the triangle 𝐴 𝐵 𝐶 will be an acute triangle if and only if, 𝐴 𝐵 + 𝐴 𝐶 > 𝐵 𝐶 . Now, suppose the triangle 𝐴 𝐵 𝐶 is not an acute triangle. So, it can be a right triangle or obtuse triangle. At first, suppose 𝐴 𝐵 𝐶 is a right triangle. The triangle 𝐴 𝐵 𝐶 will be a right triangle if and only if, 𝐴 𝐵 +𝐴 𝐶 =𝐵 𝐶 ⇒ 𝐴𝐵 (1 + 𝛽 + 𝛼 ) {2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 )} =0 (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 )(2𝛽 + 2𝛼 − 1) As we have proved that (2 + 2𝛽 − 𝛼 ), (2 + 2𝛼 − 𝛽 ) , (2𝛽 + 2𝛼 − 1) and 𝐴𝐵 (1 + 𝛽 + 𝛼 ) are greater than 0 when 1< 𝛼 < 𝛽. That’s why, 𝐴 𝐵 + 𝐴 𝐶 = 𝐵 𝐶 , this will be true if, 2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 ) = 0 ⇒ 𝛼 = 𝛽 − 4𝛽 + 1 ⇒ 𝛽 − 2𝛽 + 1 = 𝛽 − 4𝛽 + 1 [𝛼 = 𝛽 − 1] But this is impossible. In this case 𝛽 > 1, so 𝛽 − 2𝛽 + 1 ≠ 𝛽 − 4𝛽 + 1. That’s why, 𝐴 𝐵 + 𝐴 𝐶 ≠ 𝐵 𝐶 . Now suppose, the triangle 𝐴 𝐵 𝐶 is an obtuse triangle. The triangle 𝐴 𝐵 𝐶 will be an obtuse triangle if and only if, 𝐴 𝐵 +𝐴 𝐶 <𝐵 𝐶 ⇒ 𝐴𝐵 (1 + 𝛽 + 𝛼 ) {2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 )} <0 (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 )(2𝛽 + 2𝛼 − 1) Here, 𝐴 𝐵 +𝐴 𝐶 < 𝐵 𝐶 , this will be true if and only if, 21 2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 ) < 0 ⇒ 𝛼 < 𝛽 − 4𝛽 + 1 ⇒ 𝛽 − 2𝛽 + 1 < 𝛽 − 4𝛽 + 1 [𝛼 = 𝛽 − 1] But this is impossible because 𝛽 is always positive. So, 𝐴 𝐵 + 𝐴 𝐶 ≮ 𝐵 𝐶 . Now, the triangle 𝐴 𝐵 𝐶 cannot be a right triangle or obtuse triangle. So it must be an acute triangle. Theorem 1.6: If the medians of an obtuse triangle inscribed in a circle are extended, they will intersect at three points of the circumference of the circle. The triangle obtained by connecting the three points of intersection will be acute or right or obtuse. Figure 7: 𝐴 𝐵 𝐶 is an obtuse triangle inscribed in the circle. After extending the medians of the triangle 𝐴 𝐵 𝐶 , they intersect at three points of the circumference of the circle. Connecting those three points of intersection, the triangle 𝐴 𝐵 𝐶 has been obtained. As the triangle 𝐴 𝐵 𝐶 is an obtuse triangle, so it can be an isosceles triangle or scalene triangle. If 𝐴 𝐵 𝐶 is an isosceles triangle, then the triangle 𝐴 𝐵 𝐶 will be an acute isosceles triangle according to theorem 1.2. Now we have to prove that if 𝐴 𝐵 𝐶 is an scalene triangle, then the triangle 𝐴 𝐵 𝐶 will be an acute triangle. Suppose, in case of triangle 𝐴 𝐵 𝐶 , 𝐴 𝐵 < 𝐵 𝐶 < 𝐴 𝐶 , the main side = 𝐴 𝐵 , secondary side = 𝐵 𝐶 = 𝐴 𝐵 𝛼 and secondary side = 𝐴 𝐶 = 𝐴 𝐵 𝛽 . Here, 𝛼 = 𝐵 𝐶 ⁄𝐴 𝐵 and 𝐴 𝐵 < 𝐵 𝐶 , that’s why, 𝛼 > 1 . Again, 𝛽 = 𝐴 𝐶 ⁄𝐴 𝐵 and 𝐴 𝐵 < 𝐴 𝐶 , that’s why, 𝛽 is also greater than 1. Again, 𝐵 𝐶 < 𝐴 𝐶 , that’s why, 𝛼 < 𝛽 . So, we can write, 1<𝛼<𝛽 ⇒1<𝛼 <𝛽 Again, we can write (look at the proof of theorem 1.1), 𝛽 − 1 < 𝛼 < 𝛽 ⇒ (𝛽 − 1) < 𝛼 < 𝛽 Now, we can see, when 𝛽 > 1 and 𝛽 ≤ 2 , that time (𝛽 − 1) ≤ 1 and when 𝛽 > 2, that time (𝛽 − 1) > 1. As the triangle 𝐴 𝐵 𝐶 is an obtuse triangle, so 𝐴 𝐵 +𝐵 𝐶 <𝐴 𝐶 22 ⇒𝐴 𝐵 +𝐴 𝐵 𝛼 <𝐴 𝐵 𝛽 ⇒ 𝛼 < 𝛽 − 2𝛽 + 1 Here we can see 𝛽 − 2𝛽 + 1 is always less than 𝛽 because 𝛽 > 1. So we can write, (𝛽 − 1) ≤ 1 < 𝛼 < 𝛽 − 2𝛽 + 1 < 𝛽 when 1 < 𝛽 ≤ 2 ⇒ 1 < 𝛼 < 𝛽 − 2𝛽 + 1 Again, 1 < (𝛽 − 1) < 𝛼 < 𝛽 − 2𝛽 + 1 < 𝛽 when 𝛽 > 2 ⇒ (𝛽 − 1) < 𝛼 < 𝛽 − 2𝛽 + 1 Now, by using the formulas to calculate the length of the sides of the second triangle we can write, 𝐴 𝐵 = 𝐵 𝐶 = 𝐴 𝐶 = 𝐴 𝐵 (1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 ) 𝐴 𝐵 𝛼(1 + 𝛽 + 𝛼 ) (2𝛽 + 2𝛼 − 1)(2 + 2𝛼 − 𝛽 ) 𝐴 𝐵 𝛽(1 + 𝛽 + 𝛼 ) (2 + 2𝛽 − 𝛼 )(2𝛽 + 2𝛼 − 1) Here, we have considered the shortest side of the triangle 𝐴 𝐵 𝐶 as the main side, medium side as the secondary side and longest side as the secondary side . That’s why, 𝐴 𝐵 + 𝐵 𝐶 > 𝐴 𝐶 and 𝐵 𝐶 + 𝐴 𝐶 > 𝐴 𝐵 . So, the triangle 𝐴 𝐵 𝐶 will be a right triangle if 𝐴 𝐵 + 𝐴 𝐶 = 𝐵 𝐶 . Again, triangle 𝐴 𝐵 𝐶 will be an acute triangle if 𝐴 𝐵 + 𝐴 𝐶 > 𝐵 𝐶 and the triangle 𝐴 𝐵 𝐶 will be an obtuse triangle if 𝐴 𝐵 + 𝐴 𝐶 < 𝐵 𝐶 . Now, 𝐴 𝐵 +𝐴 𝐶 −𝐵 𝐶 = 𝐴𝐵 (1 + 𝛽 + 𝛼 ) {2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 )} (2 + 2𝛽 − 𝛼 )(2 + 2𝛼 − 𝛽 )(2𝛽 + 2𝛼 − 1) As we have proved that (2 + 2𝛽 − 𝛼 ), (2 + 2𝛼 − 𝛽 ) , (2𝛽 + 2𝛼 − 1) and 𝐴𝐵 (1 + 𝛽 + 𝛼 ) are greater than 0 when 1< 𝛼 < 𝛽. That’s why, the triangle 𝐴 𝐵 𝐶 will be a right triangle if, 2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 ) = 0 ⇒ 𝛼 = 𝛽 − 4𝛽 + 1 Again, the triangle 𝐴 𝐵 𝐶 will be an acute triangle if, 2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 ) > 0 ⇒ 𝛼 > 𝛽 − 4𝛽 + 1 Again, the triangle 𝐴 𝐵 𝐶 will be an obtuse triangle if, 2𝛽 + 2𝛼 − 1 + 𝛽 (2 + 2𝛼 − 𝛽 ) − 𝛼 (2 + 2𝛽 − 𝛼 ) < 0 ⇒ 𝛼 < 𝛽 − 4𝛽 + 1 We have calculated that 1 < 𝛼 < 𝛽 − 2𝛽 + 1 when 1 < 𝛽 ≤ 2 . 23 And (𝛽 − 1) < 𝛼 < 𝛽 − 2𝛽 + 1 when 𝛽 > 2. So, the triangle 𝐴 𝐵 𝐶 will be an acute triangle if, 1 < 𝛽 − 4𝛽 + 1 < 𝛽 − 2𝛽 + 1 when 1 < 𝛽 ≤ 2 Or, (𝛽 − 1) < 𝛽 − 4𝛽 + 1 < 𝛽 − 2𝛽 + 1 when 𝛽 > 2 Now suppose, 1 < 𝛽 − 4𝛽 + 1 < 𝛽 − 2𝛽 + 1 when 1 < 𝛽 ≤ 2 So, 1 < 𝛽 − 4𝛽 + 1 ⇒ 𝛽 − 4𝛽 > 0 Here we can see the solution of this inequality is, 𝛽 = (−∞, −2) ∪ (2, ∞) . But in this case 1 < 𝛽 ≤ 2. So, the acceptable solution of this inequality is, 𝛽 = {(−∞, −2) ∪ (2, ∞)} ∩ (1,2] =∅ So, there is no acceptable solution of this inequality. That’s why, 1 ≮ 𝛽 − 4𝛽 + 1 So, 𝛼 ≠ 𝛽 − 4𝛽 + 1 when 1 < 𝛽 ≤ 2 . That’s why, triangle 𝐴 𝐵 𝐶 cannot be a right triangle when 1 < 𝛽 ≤ 2 . Again, if the triangle 𝐴 𝐵 𝐶 is an obtuse triangle, then 𝛽 − 4𝛽 + 1 must be greater than 1 when 1 < 𝛽 ≤ 2 because in this case 𝛼 < 𝛽 − 4𝛽 + 1 . But we can see 𝛼 ≮ 𝛽 − 4𝛽 + 1 when 1 < 𝛽 ≤ 2. So, the triangle 𝐴 𝐵 𝐶 cannot be a right triangle or obtuse triangle when 1 < 𝛽 ≤ 2. That’s why, triangle 𝐴 𝐵 𝐶 will be an acute triangle when 1 < 𝛽 ≤ 2. Now suppose, So, (𝛽 − 1) < 𝛽 − 4𝛽 + 1 < 𝛽 − 2𝛽 + 1 when 𝛽 > 2 (𝛽 − 1) < 𝛽 − 4𝛽 + 1 ⇒ 𝛽 − 4𝛽 + 6𝛽 − 4𝛽 + 1 < 𝛽 − 4𝛽 + 1 ⇒ −4𝛽 + 10𝛽 − 4𝛽 < 0 Here we can see the solution of this inequality is, 𝛽 = 0, 𝛽 > 2. So, the acceptable solution of this inequality is, 1 𝛽 = 0, ∪ (2, ∞) ∩ (2, ∞) 2 = (2, ∞) ∪ (2, ∞). But in this case Again, 𝛽 − 4𝛽 + 1 < 𝛽 − 2𝛽 + 1 Here we can see this inequality is true for every positive value of 𝛽. That’s why, (𝛽 − 1) < 𝛽 − 4𝛽 + 1 < 𝛽 − 2𝛽 + 1, this is true when 𝛽 > 2. So, if we take 𝛼 = 𝛽 − 4𝛽 + 1 when 𝛽 > 2 the triangle 𝐴 𝐵 𝐶 will be a right triangle. Again, if we take (𝛽 − 1) < 𝛼 < 𝛽 − 4𝛽 + 1 when 𝛽 > 2 , the triangle 𝐴 𝐵 𝐶 will be an obtuse triangle. Again, if we take 𝛽 − 4𝛽 + 1 < 𝛼 < 𝛽 − 2𝛽 + 1 when 𝛽 > 2, the triangle 𝐴 𝐵 𝐶 will be an acute triangle. 4. Conclusion In this paper we have shown elementary proofs of the theorems. These 6 theorems are enough to describe the properties of the second triangle obtained by extending the medians of the first inscribed triangle in a circle. 24 5. Acknowledgement I want to thank my parents MD. Shah Alam and MST Sabina Yesmin for providing me with the help materials. 6. References 1. Byju’s n. d., Pythagoras Theorem, viewed 14 September 2020, http://byjus.com/maths/pythagoras-theorem/ 2. Byju’s n. d., Apollonius Theorem, viewed 14 September 2020, http://byjus.com/maths/apollonius-theorem/ 3. Byju’s n. d., Inequalities Calculator, viewed 14 September 2020, http://byjus.com/inequalities-calculator/ 4. Teachoo n. d., Theorem 7.8 – Chapter 7 Class 9 Triangles, viewed 14 September 2020, http://www.teachoo.com/4273/1042/Theorem-7.8---class-9---sum-of-two-sides-of-atriangle-is-greater-than/category/Theorems/ 5. http://amsi.org.au/teacher_modulas/Circle_Geometry.html Author’s Biography This is Salman Mahmud. I am an undergraduate student of Government Azizul Haque College, Bogura, Bangladesh (2022). My core research interest is in modern physics and pure mathematics. But I am interested in all the sectors of science. I have contributed in so many researches. I like to spend time with my parents MD. Shah Alam and MST Sabina Yesmin. In my leisure period I like to play video games.