In figure 1, D ABC is a right triangle, where? A is a right angle. We draw three squares ABfg, BCED and ACKH beside ab, BC and AC respectively. By drawing a straight line AL from point A to make it perpendicular to DE, intersect with L and intersect with M BC, it is not difficult to prove that D FBC is all equal to D ABD(S.A.S). So the area of the square ABFG = 2? Area of FBC = 2? Area of dabad = area of rectangular BMLD. Similarly, the area of square ACKH = the area of rectangular MCEL. That is, the area of BCED square, the area of ABFG square and the area of ACKH square, that is, AB2+AC2 = BC2. This proves the Pythagorean theorem.
This proof skillfully uses the relationship between congruent triangles, triangle area and rectangle area. Not only that, it explains the geometric meaning of "the sum of squares of two right angles" more specifically, that is, the square is divided into BMLD and MCEL with ML!
Another important significance of this proof lies in its origin. This proof comes from the hand of Euclid, an ancient Greek mathematician.
Euclid was born in 325 BC and died in 265 BC. He worked in Alexandria, the cultural center of ancient Greece, and finished his book Elements of Geometry. The Elements of Geometry is an epoch-making work, which integrates the mathematical knowledge of past dynasties and establishes a deductive system by axiomatic method, which has a far-reaching impact on the development of mathematics in later generations. The first volume of the book, proposition 47, records the above proof of Pythagorean theorem.
Evidence 2
Figure ii
In Figure 2, we put four right-angled triangles with the same size in a big square. Note that the light yellow part in the middle of the big square is also a square. Let the length of the hypotenuse of a right triangle be C and the other two sides be A and B, so since the area of a big square should be equal to the sum of the areas of four right triangles and the light yellow square in the middle, then we have it.
(a + b)2 = 4( 1/2 ab) + c2
Extension a2+2ab+b2 = 2ab+c2
Simplified to a2+b2 = c2
From this, we know that Pythagorean theorem holds.
Proof 2 can be considered as a very straightforward proof. The most interesting thing is that if we flip the right triangle in the diagram and spell it into Figure 3 below, we can still prove the Pythagorean theorem in a similar way as follows:
Figure 3
The area can be calculated as C2 = 4( 1/2 ab)+(b-a)2.
Extension = 2ab+b2-2ab+a2
Simplified c2 = a2+b2 (Theorem Proof)
Another important significance of Figure 3 is that this proof was first put forward by a China person! According to records, this is the Zhao Shuang of Wu during the Three Kingdoms period (that is, around the 3rd century AD). When Zhao Shuang annotated the classic Weekly Parallel Computing, he added an illustration he called "Pythagoras Square Diagram" (or "Chord Diagram"), which is the diagram in Figure 3 above.
Evidence 3
Figure 4
Figure 4 1 * * Draw two green congruent right triangles and a light yellow isosceles right triangle. It is not difficult to see that the whole picture has become a trapezoid. Using the trapezoidal area formula, we get:
1/2(a+b)(b+a)= 2( 1/2 ab)+ 1/2 C2
The expanded1/2a2+AB+1/2b2 = AB+1/2c2.
Simplified a2+b2 = c2 (Theorem Proof)
Some books praise the 30-minute proof, because this proof was written by an American president!
In 188 1, Garfield (James A. Garfield; ; 1831-1881) was elected as the 20th president of the United States. Unfortunately, he was assassinated five months after his election. As for the proof of Pythagorean theorem, he put forward it in 1876.
Personally, I don't think Proof 3 has any advantages. It is actually the same as Proof 2, except that it cuts the number in Proof 2 in half! What's more, I don't think the trapezoidal area formula is simpler than the square area formula!
In addition, if from a teacher's point of view, both Proof II and Proof III have the same shortcomings, we must reach the identity (AB) 2 = A2AB+B2. Although this identity is generally included in the curriculum of senior two, many students fail to fully grasp it. Because the above two proofs are used, students often can't understand and keep up in teaching.
Evidence 4
(a) (b) (c)
Figure 5
Proof 4 does this: As shown in Figure 5 (a), we draw a right triangle first, and then add a square on the triangle side next to the shortest right angle side, which is indicated in red for clarity. Add another square under the other right angle, which is represented by blue. Next, draw a square with the length of the hypotenuse, as shown in Figure 5 (b). We intend to prove that the sum of the areas of the red and blue squares is exactly equal to the area of the square drawn on the hypotenuse.
Note that in fig. 5 (b), when the hypotenuse square is added, some parts of red and blue are beyond the range of the hypotenuse square. Now I will show the out-of-range parts in yellow, purple and green respectively. At the same time, in the hypotenuse square, there are some parts that are not filled with color. Now, according to the method in Figure 5 (c), move the triangle out of range to the unfilled area. We found that the out-of-range part just filled the unfilled place! From this, we find that the sum of the areas of red and blue in Figure 5 (a) must be equal to the area of the hypotenuse square in Figure 5 (c). From this, we have confirmed the Pythagorean theorem.
This proof was put forward by Liu Hui, a mathematician of Wei State in the Three Kingdoms period. In the fourth year of Wei Jingyuan (AD 263), Liu Hui annotated the ancient book Nine Chapters Arithmetic. In the annotation, he drew a diagram similar to Figure 5 (b) to prove Pythagorean theorem. Because he used "green out" and "Zhu out" to represent yellow, purple and green, and "green in" and "Zhu in" to explain how to fill the blank part of the hypotenuse square, later mathematicians called this figure "green in and out". Others use the word "complementarity" to express the principle of this proof.
In history, Liu Hui is not the only one who proved Pythagorean theorem with the principle of "complementary entry and exit". For example, in India, in the Arab world and even in Europe, there have been similar proofs, but the pictures they drew may be somewhat different in appearance from those of Liu Hui. Fig. 6 below is a combination of fig. 5 (b) and fig. 5 (c). Note that I have redrawn a small square outside the triangle. Please look at Figure 6. Have we ever seen a similar figure?
Figure 6
In fact, isn't Figure 6 just Figure 1? It just draws the diagram 1 from another angle. Of course, the methods of dividing the parties are different.
By the way, there are obvious differences between the previous proof and the fourth proof. There is no calculation in the fourth proof, and the whole proof is only obtained by moving a few numbers. I don't know if you accept these "proofs" without any calculation steps, but I like these "wordless proofs" very much myself.
Figure 7
Among many kinds of "proof without words", I like two best. Figure 7 is one of them. The method is to divide a square with a larger right angle into four points by dividing it into a vertical line and a horizontal line. Then according to the color in Figure 7, fill two right-angled squares into the hypotenuse square, and the proof of the theorem can be completed.
In fact, there are many proofs of similar "puzzles", and I have no intention to record them here.
Another kind of "proof without words" can be regarded as the most ingenious and simple, and the method is as follows:
Evidence 5
(a) (b)
Figure VIII
Fig. 8 (a) is the same as fig. 2. Four right triangles are placed in a big square. Note that the area of the light yellow part in the figure is equal to c2. Now let's move the four right triangles in Figure 8 (a) to Figure 8 (b). Obviously, the sum of the areas of two light yellow squares in Figure 8 (b) should be a2+b2. However, since the big squares in (a) and (b) are the same and the four right triangles are equal, the areas of the remaining two light yellow parts should also be equal, so we get a2+b2 = c2 and prove the Pythagorean theorem.
There are many theories about the origin of this proof: some people say that it comes from an ancient math book in China; Some people think that Pythagoras made this proof that year, so he slaughtered a hundred cows to celebrate. In a word, I think this is the simplest and quickest proof among many proofs.
Don't underestimate this proof, it actually contains another meaning, which is not easy to be perceived by everyone. I will now "squeeze" the above two pictures into Figure 9:
(a) (b)
Figure 9
The light yellow part in the middle of fig. 9 (a) is a parallelogram, and its area can be obtained by the following formula: mn sin(a+b) where m and n are the hypotenuse lengths of two right-angled triangles respectively. The light yellow part in fig. 9 (b) is two rectangles, and the sum of their areas is: (m cos a)(n sin b)+(m sin a)(n cos b). Just like the above, the light yellow areas of (a) and (b) are equal, so combining the two formulas and eliminating the multiple of * * *, we get: sin(a+b) = sin a cos b+sin b cos a, which is the most important complex angle formula in trigonometry! It turns out that Pythagorean theorem and this complex angle formula come from the same proof!
In the second proof, after introducing the method of expanding (a+b)2, I put forward Zhao Shuang's "chord graph", which is a method of expanding (a-b)2. Prove 5 has a similar situation. Here we have a "proof without words" similar to (a+b) and a "proof without words" similar to (a-b). This method was developed by Indian mathematician Bascara; 1114-1185), as shown in figure 10.
(a) (b)
Figure X
Evidence 6
Figure Xi
In figure 1 1, we use CD to divide the middle right triangle ABC into two parts, among which? C is a right angle, D is above AB, and CD is above AB. Let a = CB, b = AC, c = AB, x = BD, y = AD. Note that the three triangles in the figure are all similar to each other, D DBC ~ D CBA ~ D DCA, so
= and =
Therefore, a2 = cx and b2 = cy are obtained.
Combining the two formulas, we get a2+b2 = cx+cy = c(x+y) = c2. Theorem proof.
Proof 6 can be said to be very special because it is the only proof in this paper that does not use the concept of area. I believe that in some old textbooks, Proof Six was also used as proof of Pythagorean theorem. However, because this proof requires the concept of similar triangles, and the two triangles have to be tossed and turned, which is quite complicated, it is rarely used in today's textbooks, and it seems to have been gradually forgotten by people!
However, if you think about it carefully, you will find that this proof is actually no different from the proof 1 (Euclid's proof)! Although this proof does not mention the area, a2 = cx actually means that the area of the square on BC is equal to the area of the rectangle formed by AB and BD, that is, the yellow part in figure 1. Similarly, b2 = cy is the dark green part in the diagram 1. From this perspective, the two proofs are based on the same principle!
Evidence 7
(a) (b) (c)
Figure XII
In figure 12 (a), we don't know the direct relationship between the areas of three squares for the time being, but because the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding sides, and any square is similar, we know that area I: area II: area III = a2: b2: c2.
But if you think about it carefully, you will find that the requirement of "Fang" in the above inference is redundant. In fact, as long as they are similar figures, such as the semicircle in figure 12 (b) or the odd shape in figure 12 (c), as long as they are similar to each other, then the area I: area II: area III is equal to a2: b2: c2!
Among many similar figures, the most useful one is a right triangle similar to the original triangle.