《一》 幾何學題篇引言

《二》 幾何學的基本邏輯概念爻

《三》 天符經

《四》 自然數是三個三角形數之和

《五》 什麼是四象 五行八卦九疇

《六》 中國算盤和九九邏輯圖

《七》 通天连列图

《八》 四易原位結構圖

《九》 九图中國文化的經典的結構1

《十》 中國文化的經典的結構2

《十一》 中國文化的經典的結構3

《十二》 中國文化的經典的結構4

《十三》 中國文化的經典的結構5

**《勾股定理的N種證明3-6》**

Proof #3Now we start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c. The square has a square hole with the side (a - b). Summing up its area (a - b) ^{2} and 2ab, the area of the four triangles (4•ab/2), we getc ^{2} = (a - b)^{2} + 2ab
= a^{2} - 2ab + b^{2} + 2ab
= a^{2} + b^{2
}
Proof #4The fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a + b) and a hole with the side c. We can compute the area of the big square in two ways. Thus(a + b) ^{2} = 4•ab/2 + c^{2
} simplifying which we get the needed identity. A proof wich combines this with proof #3 is credited to the 12th century Hindu mathematician Bhaskara (Bhaskara II): Here we add the two identities c ^{2} = (a - b)^{2} + 4•ab/2 and
c^{2} = (a + b)^{2} - 4•ab/2
which gives 2c ^{2} = 2a^{2} + 2b^{2}. The latter needs only be divided by 2. This is the algebraic proof # 36 in Loomis' collection. Its variant, specifically applied to the 3-4-5 triangle, has featured in the Chinese classic Chou Pei Suan Ching 《周髀算经》 dated somewhere between 300 BC and 200 AD and which Loomis refers to as proof 253. Proof #5This proof, discovered by President J.A. Garfield in 1876 [Pappas], is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a + b)/2•(a + b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c•c/2. As before, simplifications yield a^{2
}+ b^{2} = c^{2}.
Two copies of the same trapezoid can be combined in two ways by attaching them along the slanted side of the trapezoid. One leads to the proof #4, the other to proof #52. Proof #6We start with the original triangle, now denoted ABC, and need only one additional construct - the altitude AD. The triangles ABC, BDA and ADC are similar which leads to two ratios: AB/BC = BD/AB and AC/BC = DC/AC. Summing up we get AB•AB + AC•AC = BD•BC + DC•BC = (BD+DC)•BC = BC•BC. In a little different form, this proof appeared in the Mathematics Magazine, 33 (March, 1950), p. 210, in the Mathematical Quickies section, see Mathematical Quickies, by C. W. Trigg. Taking AB = a, AC = b, BC = c and denoting BD = x, we obtain as above a ^{2} = cx and b^{2} = c(c - x), which perhaps more transparently leads to the same identity. In a private correspondence, Dr. France Dacar, Ljubljana, Slovenia, has suggested that the diagram on the right may serve two purposes. First, it gives an additional graphical representation to the present proof #6. In addition, it highlights the relation of the latter to proof #1. It must be mentioned that this proof is just a variant of the next one - Euclid's second and less known proof of the Pythagorean proposition. Source: http://www.cut-the-knot.org/pythagoras/index.shtml |