Make four congruent right-angled triangles, and let their two right-angled sides be A and B? The length of the hypotenuse is c? Combine them into a polygon as shown in the figure, so that D, E and F are in a straight line. A little C as an extension of AC, a little DF of P?

∵? D, e and f are in a straight line. And rt δ GEF? ≌? rtδEBD，？

∴? ∠EGF？ =? ∠ Bed?

∵? ∠EGF？ +? ∠ Global Environment Facility? =? 90 ,?

∴? ∠ bed +? ∠ Global Environment Facility? =? 90 ,?

∴? ∠ beg? = 180 ―90 =? 90 ?

Again? AB？ =? Is it? =? EG？ =? Hey? =? c，？

∴? ABEG is a square with side length c?

∴? ∠ABC？ +? ∠CBE？ =? 90 ?

∵? rtδABC？ ≌? rtδEBD，？

∴? ∠ABC？ =? ∠EBD。 ?

∴? ∠EBD？ +? ∠CBE？ =? 90 ?

Namely. ∠CBD=？ 90 ?

Again? ∠BDE？ =? 90，∠BCP？ =? 90 ,?

BC? =? BD？ =? Answer?

∴? BDPC is a ...?

Similarly, HPFG is a square with a side length of B.

Let the area of polygon GHCBE be s, then?

A2+B2=C2？

Proof method 2

Make two congruent right-angled triangles, and let their two right-angled sides be A and B (B >; a)？ The length of the hypotenuse is c? Make a square with a side length of c, and put them together to form a polygon as shown in the figure, so that E, A and C are in a straight line.

Q is QP∑BC, and AC is at point P?

Point B is BM⊥PQ, and the vertical foot is m; A little more?

F is FN⊥PQ, and the vertical foot is n?

∵? ∠BCA？ =? 90, QP∨ BC,?

∴? ∠MPC？ =? 90 ,?

∵? BM⊥PQ？

∴? ∠BMP？ =? 90 ,?

∴? BCPM is a rectangle, namely ∠MBC? =? 90 。 ?

∵? ∠QBM？ +? ∠MBA？ =? ∠QBA？ =? 90 ,?

∠ABC？ +? ∠MBA？ =? ∠MBC？ =? 90 ,?

∴? ∠QBM？ =? ∠ABC，？

Again? ∠BMP？ =? 90，∠BCA？ =? 90，BQ？ =? Ba? =? c，？

∴? rtδBMQ？ ≌? rtδBCA。 ?

Similarly, rt δ qnf can prove it? ≌? Rt δ AEF。 That is A2+B2=C2?

Proof method 3

Make two congruent right-angled triangles, and let their two right-angled sides be A and B (B >; a)？ The length of the hypotenuse is c? Make another square with a side length of c and put them together into a polygon as shown in the figure.

CF and AE are used as squares with sides of FCJI and AEIG 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 ,?

∴rtδcjb？ ≌? rtδCFD？ ,?

Similarly, rt δ abg? ≌? rtδADE，？

∴rtδcjb？ ≌? rtδCFD？ ≌? rtδABG？ ≌? rtδADE？

∴∠ABG？ =? ∠BCJ，？

∵∠BCJ？ +∠CBJ=？ 90 ,?

∴∠ABG？ +∠CBJ=？ 90 ,?

∵∠ABC=？ 90 ,?

G, B, I and J are on the same line.

A2+B2=C2 .？

Proof method 4

Make a triangle with three sides A, B and C, and put it in the shape as shown in the figure, so that the three points H, C and B are in a straight line and connected?

BF、CD。 ? C is CL⊥DE,

AB crosses at m and DE crosses at l.

∵? AF？ =? AC，AB？ =? AD，？

∠FAB？ =? ∠GAD，？

∴? FAB？ ≌? GAD，？

∵? Is the area of δδFAB equal to?

The area of GAD is equal to the right angle ADLM?

Half the area?

∴? The area of rectangular ADLM? =.？

Similarly, what is the area of the rectangular MLEB? =.？

∵? The area of a square ADEB?

=? The area of rectangular ADLM? +? The area of the rectangular MLEB?

∴? That is A2+B2=C2?

Proof 5 (Euclid Proof)

Proof in the Elements of Geometry?

In Euclid's Elements of Geometry, the Pythagorean theorem was proved as follows. 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 angles between the two sets of sides are 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 evidence 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 then what? g？ They are all linear correspondences, and B, A and H can all be proved in the same way. ∠CBD and∠ ∠FBA are right angles, so∠ ∠ABD is equal to∠ ∠FBC. Because? AB？ And then what? BD？ Are they equal? FB？ And then what? BC, so △ABD? Must be equal to △FBC. Because? Answer? With what? k？ And then what? L is linear, so square? BDLK？ Must be twice the area of △ABD. Because c, a and g have the same linearity, the square of BAGF must be twice that of △FBC. So quadrilateral? BDLK？ Must have the same area. Buff? =? AB & ampsup2。 The same can be proved, quadrilateral? Kohler? Must have the same area. ACIH？ =? AC2； . Add up these two results. AB2； +? AC2； ; ? =? BD×BK？ +? KL×KC. Since 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, in Rt△ABC, ∠ ABC = 90, BD is the height on the hypotenuse AC?

By proving that triangles are similar, there are the following projective theorems:

( 1)(BD)2； =AD？ DC？

(2)(AB)2； =AD？ AC？ ,?

(3) (BC) 2; =CD？ Communication.

From formula (2)+(3): (ab) 2; +(BC)2； =AD？ AC+CD？ AC？ =(AD+CD)？ AC =(AC)2； ,?

Figure 1 Yes? (AB)2； +(BC)2； =(AC)2, which is the conclusion of Pythagorean theorem. ?

Figure 1

Proof 7 (Zhao Shuang's String Diagram)

In this Pythagorean Square Diagram, 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 you can get the following formula:?

4×(ab/2)+(b-a)2？ = c2？

After simplification, you can get: a2? +b2？ = c2？

Namely: c=(a2? +b2？ ) 1/2?

Another name for Pythagorean theorem? Pythagorean theorem is a dazzling pearl in geometry, which is called "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 rings are shared, it is 345. The total length of two moments is twenty plus five, which is called the product moment. "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? Zhou Kuaijing? Cultural Relics Publishing House,1March, 980, photocopied according to the six-year edition of Jiading in Song Dynasty, 1-5 pages. ?

2.？ The relationship between the proof of Zhou Pythagorean theorem and complementary principle. Published in Sinology Research 1989, Volume 7 1, pp. 255-28 1. ?

3.？ Li Guowei:? On Zhou Pian Hui Su 'an Jing and the Method of Counting from the Prescriptions. Published in Proceedings of the Second Symposium on the History of Science. Taiwan Province Province 199 1 July,? Pages 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:? The Proof of Pythagorean Theorem by Shang Gao, Zhao Shuang and Liu Hui. Published in 20 volumes of Mathematical Newsletter. September 3rd, Taiwan Province Province 1996? 20-27 pages?

Proof Method 8 (Da Vinci Proof Method)

Da Vinci's proof method

Three pieces of paper are actually the same piece of paper. After being disassembled and put together, the area of the "hole" in the middle is still the same before and after, but the expression of the area is no longer the same. If these two different expressions are equal, we can get a new relationship-Pythagorean theorem. All the proof methods of Pythagorean theorem have this in common. Observing the first piece of paper, it is easy to know EB⊥CF because we want to prove Pythagorean theorem, and because both sides of the paper are symmetrical, we can know that the quadrilateral ABOF and CDEO are both squares. Then we need to know that angle A' and angle D' are right angles. The reason can be found in the paper 1, connecting AD, because it is symmetrical, so ∠ bad = ∠ fad = ∠ CDA = ∠ EDA = 45. Obviously, in Figure 3, angle A' and angle D' are both right angles. ?

Prove:

The area of polygon ABCDEF in the first picture is S 1=S squared ABOF+S squared CDEO+2S△BCO=OF2+OE2+OF? OE？

In the third picture, the area S2=S of polygon A'B'C'D'E'F = s square B'C'E'F'+2△C'D'E'=E'F'2+C'D'? What do you mean?

Because S 1=S2?

What about 2+OE2+? OE=E'F'2+C'D？ What do you mean?

And because C'D'=CD=OE, D'E'=AF=OF?

So OF2+OE2=E'F'2?

Because E'F'=EF?

What about 2+OE2=EF2?

Proof of pythagorean theorem. ?

Proof method 9

A rectangle and three triangles can be obtained from this picture, and the derivation formula is as follows:?

b？ (? Answer? +? b？ )=? 1/2c2？ +? ab？ +? 1/2(b？ +? a)(b？ -? a)？

Rectangular area? = (middle triangle)+(bottom) 2 right triangles+(top) 1 straight?

Angle triangle. ?

(Simplified)? 2ab？ +? 2b2=? C2； ? +? B2； -? a2； +? 2ab？

2b2？ -? b2？ +? a2？ =? C2； ?

a2？ +? b2？ =? C2； ?

Note: More graphs are from Garfield diagram. ?

Proof method 10

In the Rt triangle ABC, the angle C=90 degrees, so CH is perpendicular to AB and H ...

Let a/sinA=b/sinB=c/sinC=d?

1 = sin 90 = sinC = c/d = AH/d+BH/d = cosa×b/d+cosb×a/d = cosa×sin b+cosb×sinA = a/c？ Air conditioning+air conditioning? b/c？

=(a^2+b^2)/c^2= 1？

So a 2+b 2 = c 2?

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