∫D, e, f are in a straight line, rtδGEF≌rtδEBD,
∴ ∠EGF = ∠BED，
∫∠EGF+∠GEF = 90，
∴ ∠BED + ∠GEF = 90，
∴ ∠BEG = 180 ―90 = 90
AB = BE = EG = GA = c，
Abegg is a square with a side length of C.
∴ ∠ABC + ∠CBE = 90
∫rtδABC≌rtδEBD，
∴ ∠ABC = ∠EBD。
∴∠ EBD+∠ CBE = 90, that is ∠ CBD = 90.
∠∠BDE = 90，∠ BCP = 90，BC = BD = A .
∴ BDPC is a square with a side length of 100.
Similarly, HPFG is a square with a side length of b, let the area of polygon GHCBE be s, then make two congruent right-angled triangles, and let their two right-angled sides be a and b (b >; A) Make a square with a side length of c and a hypotenuse length of c, and then put them together to make a polygon as shown in the figure, so that E, A and C are in a straight line. The intersection q is qp∨BC, the intersection AC is at point P, the intersection B is BM⊥PQ, and the vertical foot is m; F is FN⊥PQ, and the vertical foot is n.
∫∠BCA = 90, QP∨ BC.
∴ ∠MPC = 90，
* bm⊥pq，
∴ ∠BMP = 90，
∴ BCPM is a rectangle, that is ∠ MBC = 90.
∠∠QBM+∠MBA =∠QBA = 90，∠ABC + ∠MBA = ∠MBC = 90，
∴ ∠,
∵∠ BMP = 90，∠ BCA = 90，BQ = BA = c，
∴rtδbmq≌rtδBCA。 In the same way, we can prove rtδqnfδrtδAEF. That is, make two congruent right-angled triangles, and make a square with side length c by the same proof method 2. Combine them into polygons as shown in the figure. Make square FCJI and AEIG with CF and AE as side lengths respectively.
∫EF = DF-DE = b-a，EI=b，
∴FI=a，
∴ g, I and j are on the same line,
CJ = CF = a，CB=CD=c，∠CJB = ∠CFD = 90，
∴ rt δ cjb ≌ rt δ CFD, in the same way, rt δ abg ≌ rt δ ade,
∴rtδcjb≌rtδCFD≌rtδabg≌rtδade
∴∠ABG = ∠BCJ，
∠∠BCJ+∠CBJ = 90，
∴∠ABG +∠CBJ= 90，
∫∠ABC = 90 °,
G, B, I and J are on the same line. Make three triangles with side lengths of A, B and C, and put them together as shown in the figure, so that H, C and B are in a straight line, connecting BF and CD. The intersection point C is CL⊥DE, AB intersects at point M and DE intersects at point L. 。
AF = AC，AB = AD，∠FAB = ∠GAD，
∴δfab≌δgad，
∫δFAB has an area equal to, GAD has an area equal to half the area of rectangular ADLM,
∴ The area of rectangle ADLM =. Similarly, the area of rectangular MLEB =.
Area of squared ADEB = area of rectangular ADLM+area of rectangular MLEB.
∴ That is, the proof put forward in the Elements of Geometry. After Euclid thought that Pythagorean theorem could be established in the Elements of Geometry, the following proof was made. Let △ABC be a right triangle, where A is a right angle. Draw a straight line from point A to the opposite side so that it is perpendicular to the opposite square. This line divides the opposite square in two, and its area is equal to the other two squares. In formal proof, we need the following four auxiliary theorems: If two triangles have two sets of corresponding sides and the included angle between the two sets of sides is equal, then the two triangles are congruent. (SAS Theorem) The area of a triangle is half that of any parallelogram with the same base and height. The area of any square is equal to the product of its sides. The area of any square is equal to the product of its two sides (according to Auxiliary Theorem 3). The concept of proof is: transform the upper two squares into two parallelograms with equal areas, and then rotate and transform them into the lower two rectangles with equal areas. The proof is as follows: Let △ABC be a right triangle, and its right angle is CAB. Its sides are BC, AB and CA, which are drawn into four squares in turn: CBDE, Baff and ACIH. Draw parallel lines where BD and CE intersect with point A, and this line will intersect BC and DE at right angles at points K and L respectively. Connect CF and AD respectively to form two triangles BCF and BDA. ∠CAB and ∠BAG are right angles, so C, A and G are all linear correspondences, and B, A and H can also be proved in the same way. ∠CBD and∠ ∠FBA are right angles, so∠ ∠ABD is equal to∠ ∠FBC. Because AB and BD are equal to FB and BC respectively, △ABD must be equal to △FBC. Because a corresponds linearly to k and l, the square of BDLK must be twice that of △ABD. Because c, a and g are collinear, the square of BAGF must be twice the area of △FBC. So the quadrilateral BDLK must have the same area BAFF = AB &；; sup2。 Similarly, quadrilateral infilled walls must have the same area ACIH = AC2；; ; . Add these two results, AB2；; +AC2； ; = BD×BK+KL×KC. Because BD=KL, BD×BK+KL×KC = BD(BK+KC) = BD×BC Because CBDE is a square, AB2；; +AC2； = BC2； . This proof was put forward in section 1.47 of Euclid's Elements of Geometry.
Proof 6 (Proof of Euclid's Projective Theorem)
As shown in figure 1, Rt△ABC, ∠ ABC = 90, BD is the Qualcomm on the hypotenuse AC. To prove that triangles are similar, there is a projective theorem as follows:
⑴(BD)2； = ad DC,
⑵(AB)2； =AD AC，
(3) (BC) 2; =CD AC. From formula (2)+(3): (ab) 2; +(BC)2； = AD AC+CD AC =(AD+CD)AC =(AC)2； , figure 1 is (ab) 2; +(BC)2； =(AC)2, which is the conclusion of Pythagorean theorem. Figure 1 In this Pythagorean Square Figure, the square ABDE with the chord as the side length is composed of four equal right triangles and a small square in the middle. The area of each right triangle is AB/2; If the side length of a small square is b-a, the area is (b-a)2. Then the following formula can be obtained: 4× (AB/2)+(B-A) 2; = c2 can be simplified to: A2; +B2； = c2 means: c =(a2；; +B2； Pythagorean Theorem, alias Pythagorean Theorem 1/2, is a dazzling pearl in geometry, known as "the cornerstone of geometry", and is also widely used in higher mathematics and other disciplines. Because of this, several ancient civilizations in the world have been discovered and widely studied, so there are many names. China is one of the earliest countries to discover and study Pythagorean theorem. Ancient mathematicians in China called the right triangle pythagorean, the short side of the right angle is called hook, the long side of the right angle is called strand, and the hypotenuse is called chord, so the pythagorean theorem is also called pythagorean chord theorem. BC 1000 years, according to records, Shang Gao (about BC 1 120) replied to the Duke of Zhou, "Therefore, the moment is folded, thinking that the sentence is three in width, four in stock and five in diameter." If the square is square, the outer half is moment, and the ring is * * *, that is 345. Two moments * * * are twenty and five, which are called product moments. "Therefore, Pythagorean theorem is also called" quotient height theorem "in China. In the 7th-6th century BC, Chinese scholar Chen Zi once gave the trilateral relationship of any right triangle, that is, "Take the sunset as the hook, the height of the sun as the strand, and multiply and divide the hook with the strand to get evil to the sky. In France and Belgium, Pythagorean Theorem is also called "Donkey Bridge Theorem". Other countries call Pythagorean Theorem "Square Theorem". One hundred and twenty years after Chen Zi, the famous Greek mathematician Pythagoras discovered this theorem, so many countries in the world called Pythagorean theorem "Pythagorean theorem". To celebrate the discovery of this theorem, the Pythagorean school killed 100 cows as a reward for offering sacrifices to the gods, so this theorem is also called the "hundred cows theorem". Garfield, 20th President of the United States, proved Pythagorean Theorem (1876 April 1).
1 Weekly Ji Jing, Cultural Relics Publishing House, March 1980, photocopied according to the six-year edition of Jiading in Song Dynasty, 1-5 pages.
2. The relationship: the proof of Zhou's Pythagorean theorem, the entry and exit of parallel computing and complementary principle. Published in Sinology Research, Volume 7,No. 1,No. 1989, pp. 255-28 1.
3. Li Guowei: On the Zhou suan Jing and the chapter "The method of saying that the number is high comes from one side". Published in Proceedings of the Second Symposium on the History of Science, Taiwan Province Province, June 199 1, pp. 227-234.
4. Min: the differential of quotient height theorem. Published in Research on the History of Natural Science, Vol. 1993,No. 12, No.29-4 1.
5. Qu Anjing: Proof of Pythagorean Theorem by Shang Gao, Zhao Shuang and Liu Hui. Published in Mathematical Communication, Volume 20, Taiwan Province Province, No.3, 65438+,1September 1996, pp.20-27: Da Vinci's Proof Method.
It is proved that the area of polygon ABCDEF in the first picture is S 1=S squared ABOF+S squared cdeo+2s△ BCO = of2+oe2+of OE; The area of polygon A'B'C'D'E'F in the third picture is S2=S squared b' c' e' f'+2 △ C.
so of 2+oe2+of OE = e ' f ' 2+c ' d ' e '
And because C'D'=CD=OE, D'E'=AF=OF.
So OF2+OE2=E'F'2
Because E'F'=EF
Thus, the Pythagorean theorem of 2+OE2=EF2 is proved. You can get a rectangle and three triangles from this picture, and the derivation formula is as follows:
b(a+b)= 1/2 C2； +ab+ 1/2(b+a)(b-a) Rectangular area = (middle triangle)+(bottom) 2 right triangles+(top) 1 right triangle. (Simplified) 2ab+2 B2；; = c2+B2； -a2； +2ab 2b 2； -B2； +a2； = c2a2； +B2； = c2
Note: More graphs are from Garfield diagram.