Mathematics knowledge of the second day of junior high school .. Pythagorean theorem

In China, the characteristic that the sum of squares of two right angles of a right triangle is equal to the square of the hypotenuse is called Pythagorean Theorem or Pythagorean Theorem.

Theorem:

If the two right angles of a right triangle are A and B and the hypotenuse is C, then A? +b？ =c？ ; That is to say, the sum of squares of two right angles of a right triangle is equal to the square of the hypotenuse.

If the three sides A, B and C of a triangle satisfy A 2+B 2 = C 2, for example, a right-angled side is 3, a right-angled side is 4, and the hypotenuse is 3*3+4*4=X*X, and X=5. Then this triangle is a right triangle. (called the inverse theorem of Pythagorean theorem)

Source:

Pythagoras tree Pythagoras tree is a basic geometric theorem, which is traditionally proved by Pythagoras in ancient Greece. It is said that after Pythagoras proved this theorem, he beheaded a hundred cows to celebrate, so it is also called "Hundred Cows Theorem". In China, Zhou Kuai Shu Jing recorded a special case of Pythagorean Theorem, which was reportedly discovered by Shang Dynasty's Shang Dynasty, so it is also called Shang Gao Theorem. Zhao Shuang in the Three Kingdoms period made detailed comments on Pythagorean Theorem as proof in Zhou Bi suan Jing. France and Belgium are called donkey bridge theorem, and Egypt is called Egyptian triangle. In ancient China, the shorter right angle side of a right triangle was called a hook, the longer right angle side was called a chord, and the hypotenuse was called a chord.

Books on Pythagorean Theorem

Principles of Mathematics People's Education Press

Exploring Pythagorean Theorem Tongji University Press

Peking University Press: You Yinpei teaches mathematics.

Pythagoras model new century publishing house

Book nine chapters arithmetic

You Yinpei Reveals Pythagorean Theorem Jiangxi Education Publishing House

[Edit this paragraph] The earliest Pythagorean theorem

According to many clay tablets, Babylonians were the first people in the world to discover Pythagoras theorem. This is just an example. For example, in BC 1700, the ninth question on a clay tablet (No.BM85 196) was to the effect that "there is a wooden beam (AB) with a length of 5 meters vertically leaning against the wall, and the upper end (A) slides down one meter to d .. How far is the lower end (C) from the wall root (B)?" They solved the problem with Pythagorean theorem, as shown in the figure.

Let AB = CD = L = 5m, BC=a, AD = H = 1m, BD = L-H = 5- 1m = 4m.

∴ A = √ [L-(L-H)] = √ [5-(5-1)] = 3m, ∴ Triangle BDC is a twisted triangle with 3, 4 and 5 sides.

[Edit this paragraph] Introduction to Zhou Kuai Shu Jing

Green-Zhu visits the map

Zhou Kuai Kuai Jing is one of the ten books of calculation. Written in the second century BC, it was originally named Zhou Jie, which is the oldest astronomical work in China. It mainly expounded the theory of covering the sky and the method of four seasons calendar at that time. In the early Tang Dynasty, it was stipulated as one of imperial academy's teaching materials, so it was renamed Zhou Kuai. The main achievement of Zhouyi ·suan Jing in mathematics is the introduction of Pythagorean theorem and its application in measurement. The original book did not prove Pythagorean theorem, but the proof was given by Zhao Shuang in Zhou Zhuan Pythagorean Notes. ·suan Jing of Zhouyi adopts quite complicated fractional algorithm and Kaiping method. The Pythagorean theorem says: "The number method is based on a square, the square is based on a moment, and the distance is based on 998 1, so the moment is folded. The relationship of a right triangle is that the sum of squares of two right-angled sides is equal to the square of the hypotenuse, (a*a)+(b*b)=(c*c)"

The triangle is a right triangle, the square with hook A as the side is Zhu Fang, and the square with chain B as the side is square. Make up for the deficiency with surplus, and merge Zhu Fang and Fang Qing into a Li Fang. According to its area relation, there is a+b = C, because Zhu Fang and Fang Qing are on the metaphysical side, and that part will not move.

The square with the hook as the edge is Zhu Fang, and the square with the rope as the edge is Fang Qing. To make up for the deficiency, just move Zhu Fang's I(a2) to I', Fang Qing's II to II' and III to III', and you will spell out a square (c2). Take the chord as the side length. It can be proved that a2+b2=c2.

[Edit this paragraph] Garfield's story of proving Pythagorean theorem

1876 One weekend evening, on the outskirts of Washington, D.C., a middle-aged man was walking and enjoying the beautiful scenery in the evening. He was Ohio and party member Garfield. Walking, he suddenly found two children talking about something with rapt attention on a small stone bench nearby, arguing loudly and discussing in a low voice. Driven by curiosity, Garfield followed the sound and came to the two children to find out what they were doing. I saw a little boy bend down and draw a right triangle on the ground with branches. So Garfield asked them what they were doing. The little boy said without looking up, "Excuse me, sir, if the two right angles of a right triangle are 3 and 4 respectively, what is the length of the hypotenuse?" Garfield replied, "It's five." The little boy asked again, "If the two right angles are 5 and 7 respectively, what is the length of the hypotenuse of this right triangle?" Garfield replied without thinking, "The square of the hypotenuse must be equal to the square of 5 plus the square of 7." The little boy said, "Sir, can you tell me the truth?" Garfield was speechless, unable to explain, and very unhappy. Garfield stopped walking and immediately went home to discuss the questions the little boy gave him. After repeated thinking and calculation, he finally figured it out and gave a concise proof method.

As follows:

Solution: In the grid, the sum of the areas of two small squares with right angles is equal to the areas of two squares with hypotenuse.

The content of Pythagorean theorem: the sum of squares of two right angles A and B of a right triangle is equal to the square of hypotenuse C,

a^2+b^2=c^2

Explanation: Ancient scholars in China called the shorter right-angled side of right-angled triangle "hook", the longer right-angled side "chord" and the hypotenuse "chord", so they called this theorem "Pythagorean Theorem". Pythagorean theorem reveals the relationship between the sides of a right triangle.

For example, if the two right angles of a right triangle are 3 and 4 respectively, then the hypotenuse C 2 = A 2+B 2 = 9+ 16 = 25, that is, c=5.

Then the hypotenuse is 5.

Three Proof Methods of Pythagorean Theorem (Ⅱ)

Proof method 1 (Mei Wending proof)

Make four congruent right-angled triangles, let their two right-angled sides be A and B respectively, and the hypotenuse be C, and make them into polygons as shown in the figure, so that D, E and F are in a straight line. As an extension of AC passing through C, it intersects DF at point P. 。

∫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，

Abeg 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 again? ，∠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

,

∴ .

Proof Method 2 (Proof to Minda)

Make two congruent right-angled triangles, and let their two right-angled sides be A and B (B >; A), the hypotenuse length is C. Then make a square with a side length of C, and make them into polygons as shown in the figure, so that E, A and C are in a straight line.

Pass point q as QP BC, and intersect AC 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, that is ∠MBC = 90? .

∠∠QBM+∠MBA =∠QBA = 90？ ,

∠ABC + ∠MBA = ∠MBC = 90？ ,

∴ ∠QBM = ∠ABC，

∵ ∠BMP = 90 again? ，∠BCA = 90？ ，BQ = BA = c，

∴rtδbmq≌rtδBCA。

Similarly, rt δ qnf ≌ rt δ AEF can also be proved.

Proof Method 3 (Zhao Haojie Proof)

Make two congruent right-angled triangles, and let their two right-angled sides be A and B (B >; A), the hypotenuse is C. Then make a square with a side length of C and put them together to form a polygon 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？ ,

∴rtδcjb rtδCFD，

Similarly, 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,

Proof 4 (Euclid Proof)

Make three squares with sides a, b and c, and put them in the shape as shown in the figure, so that H, c and b are connected into a straight line.

BF，CD。 Exceeding c as CL⊥DE,

AB intersects at point m, and DE intersects at point.

length

AF = AC，AB = AD，

∠FAB = ∠GAD，

∴δfab≌δgad，

∫δFAB has an area equal to,

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

Half the area,

The ∴ area of rectangle ADLM =.

Similarly, the area of rectangular MLEB =.

The area of ADEB square

= area of rectangular ADLM+area of rectangular MLEB

That is ∴.

[Edit this paragraph] The alias of 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 the first country 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 years) replied that "Gou Guang San, Gu, Wu" means "Gou San, Gu Si" is a right triangle. So Pythagorean theorem is also called "high" in China.

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's death, the famous Greek mathematician Pythagoras discovered this theorem, so many countries in the world called it Pythagoras theorem. In order to celebrate the discovery of this theorem, the Pythagorean school killed one hundred cows as a reward for offering sacrifices to the gods, so this theorem is also called the "Hundred Cows Theorem".

[Edit this paragraph] Proof

There are many ways to prove this theorem, and the method to prove it may be the most among many theorems in mathematics. Elisha Scott Loomis's Pythagorean proposition always mentions 367 ways of proof.

Some people will try to prove Pythagorean theorem by trigonometric identities (such as Taylor series of sine and cosine functions), but all basic trigonometric identities are based on Pythagorean theorem, so they cannot be used as proof of Pythagorean theorem (see circular argument).

Using similar triangles's proof method

Using similar triangles's proof

There are many ways to prove Pythagorean theorem, all of which are based on the ratio of the lengths of two sides of similar triangles.

Let ABC be a right triangle, and the right angle is at the angle C (see the attached figure). Draw the height of the triangle from point C and call it the intersection of this height and AB H. This new triangle ACH is similar to the original triangle ABC, because both triangles have a right angle (because of the definition of "height") and both triangles have the same angle A, so we can know that the third angle is equal. Similarly, triangle CBH and triangle ABC are similar. These similar relationships derive the following ratio relationships:

Because BC = A, AC = B, AB = C.

So a/c = HB/a and b/c = ah/b.

It can be written as a*a=c*HB and b*b=C*AH.

Combining these two equations, we get a * a+b * b = c * HB+c * ah = c * (HB+ah) = c * C.

In other words: A * A+B * B = C * C

[*]-is a multiplication symbol.

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.

This is proved 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 BAGF = AB? . Similarly, the areas of quadrangles must be equal. ACIH = AC? . Add up these two results, AB? + AC？ = BD×BK+KL×KC Since BD=KL, BD×BK+KL×KC = BD(BK+KC) = BD×BC Since CBDE is a square, AB? + AC？ = C？ . This proof was put forward in section 1.47 of Euclid's Elements of Geometry.