"商高定理"的搜尋報告

勾股定理，又称畢達哥拉斯定理或畢氏定理。是一個基本的幾何定理，傳統上認為是由古希腊的畢達哥拉斯所證明。據說畢達哥拉斯證明了這個定理後，即斬了百頭牛作慶祝，因此又稱「百牛定理」。在中國，《周髀算經》記載了勾股定理的一個特例，相传是在商代由商高發現，故又有稱之為商高定理；三国时代的赵爽对《周髀算經》内的勾股定理作出了详细注释，作為一個證明。法国和比利时称为驴桥定理，埃及称为埃及三角形。

定理



勾股数组是滿足勾股定理a)，其中。

勾股數组
這個定理的歷史可以被分成三個部份:發現畢氏三元數(勾股數)、發現直角三角形中邊長的關係、定理的證明 knowledge of Pythagorean triples, knowledge of the relationship between the sides of a right triangle, and proofs of the theorem.
Megalithic monuments from circa 2500 BC in Egypt, and in the British Isles, incorporate right triangles with integer sides.
有許多辯論是否畢氏定理早已不只一次被發現There is much debate on whether the Pythagorean theorem was discovered once or many times. B.L. van der Waerden asserts a single discovery, 有人說是西元前2000年從英國發現，然後傳播到達米亞by someone in Neolithic Britain, knowledge of which then spread to Mesopotamia circa 2000 BC, and from there to India, China, and Greece by 600 BC. 然而許多學者並不同意這 種說法Most scholars disagree however, and favor independent discovery.
最近，Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaja在吠陀數學一書中聲稱古代印度教吠陀證明了畢達哥拉斯定理。More recently, Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaja in his book Vedic Mathematics claimed ancient Indian Hindu Vedic proofs for the Pythagoras Theorem.

歷史
這個定理有許多證明的方法，其證明的方法可能是數學眾多定理中最多的。(the law of quadratic reciprocity being also a contender for that distinction); 路明思（Elisha Scott Loomis）的 Pythagorean Proposition, 一書中總共提到 367 證明方式。
Some arguments based on trigonometric identities (such as Taylor series for sine and cosine) have been proposed as proofs for the theorem. However, since all the fundamental trigonometric identities are proved using the Pythagorean theorem, there cannot be any trigonometric proof. (See also begging the question.)

證明
有許多畢氏定理的證明方式，都是基於相似三角形中兩邊長的比例。
設ABC為一直角三角形, 直角於角C（看附圖）. 從點C畫上三角形的高，並將此高與AB的交叉點稱之為H。此新三角形ACH和原本的三角形ABC相似，因為在兩個三角形中都有一個直角（這又是由於「高」的定義），而兩個三角形都有A這個共同角，由此可知第三隻角都是相等的。同樣道理，三角形CBH和三角形ABC也是相似的。這些相似關係衍生出以下的比率關係：
因為

所以

可以寫成

綜合這兩個方程式，我們得到

換句話說:

利用相似三角形的證法
在歐幾里得的幾何原本一書中提出畢氏定理由以下証明後可成立。 設△ABC為一直角三角形，其中A為直角。Drop a perpendicular from A to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.
在正式的證明中，我們需要四個輔助定理如下：
The intuitive idea behind this proof, which can make it easier to follow, is that the top squares are morphed into parallelograms with the same size, then turned and morphed into the left and right rectangles in the lower square, again at constant area.
其證明如下：
此證明是於歐幾里得《幾何原本》一書第1.47節所提出的

如果兩個三角形有兩組對應邊和對應角相等，則兩三角形全等。
三角形面積是任一同底同高之平行四邊形面積的一半。
任意一個正方形的面積等於其二邊長的乘積。
任意一個四方形的面積等於其二邊長的乘積（據輔助定理3）。
設△ABC為一直角三角形，其直角為CBA。
其邊為BC、AB、和CA，依序繪成四方形CBDE、BAGF和ACIH。
畫出過點A之BD、CE的平行線。It will perpendicularly intersect BC and DE at K and L, respectively.
Join CF and AD, to form the triangles BCF and BDA.
∠CAB和∠BAG都是直角，因此C、A 和 G 都是線性對應的are colinear，同理可證B、A和H。
∠CBD和∠FBA皆為直角，所以∠ABD等於∠FBC。
因為 AB 和 BD 分別等於 FB 和 BC ，所以△ABD 必須相等於△FBC。
因為 A is colinear with K 和 L, 四方形 BDLK 必須二倍面積於△ABD。
因為C、A和G有共同線性，正方形BAGF必須二倍面積於△FBC。
因此四邊形 BDLK 必須有相同的面積 BAGF = AB².
同理可證，四邊形 CKLE 必須有相同的面積 ACIH = AC².
加入這兩個結果值會到, AB²+ AC² = BD×BK + KL×KC
由於BD=KL，BD×BK + KL×KC = BD(BK + KC) = BD×BC
由於CBDE是個正方形，因此AB² + AC² = C²，.

歐幾里得的証法
From the same diagram as that in Euclid's proof above, we can see three similar figures, each being "a square with a triangle on top". Since the large triangle is made of the two smaller triangles, its area is the sum of areas of the two smaller ones. By similarity, the three squares are in the same proportions relative to each other as the three triangles, and so likewise the area of the larger square is the sum of the areas of the two smaller squares.

歐幾里得的相似三角形証法
A proof by rearrangement is given by this illustration. The area of each large square is (a + b)². In both, the area of four identical triangles is removed. The remaining areas, a² + b² and c², are equal. Q.E.D.
This proof is indeed very simple, but it is not elementary, in the sense that it does not depend solely upon the most basic axioms and theorems of Euclidean geometry. In particular, while it is quite easy to give a formula for area of triangles and squares, it is not as easy to prove that the area of a square is the sum of areas of its pieces. In fact, proving the necessary properties is harder than proving the Pythagorean theorem itself (see Lebesgue measure). For this reason, axiomatic introductions to geometry usually employ another proof based on the similarity of triangles (see above).
A second graphic illustration of the Pythagorean theorem fits parts of the sides' squares into the hypotenuse's square. A related proof would show that the repositioned parts are identical with the originals and, since the sum of equals are equal, that the corresponding areas are equal. To show that a square is the result one must show that the length of the new sides equals c. Note that for this proof to work, one must provide a way to handle cutting the small square in more and more slices as the corresponding side gets smaller and smaller.

圖形重新排列證法
An algebraic variant of this proof is provided by the following reasoning. Looking at the illustration which is a large square with identical right triangles in its corners, the area of each of these four triangles is given by

The A-side angle and B-side angle of each of these triangles are complementary angles, so each of the angles of the blue area in the middle is a right angle, making this area a square with side length C. The area of this square is C.
(Distribution of the 4)
(Subtraction of 2AB)

代數證法
One can arrive at the Pythagorean theorem by studying how changes in a side produce a change in the hypotenuse in the following diagram and employing a little calculus.
As a result of a change in side a,

by similar triangles and for differential changes. So

upon separation of variables. A more general result is

which results from adding a second term for changes in side b.
Integrating gives


So

As can be seen, the squares are due to the particular proportion between the changes and the sides while the sum is a result of the independent contributions of the changes in the sides which is not evident from the geometric proofs. From the proportion given it can be shown that the changes in the sides are inversely proportional to the sides. The differential equation suggests that the theorem is due to relative changes and its derivation is nearly equivalent to computing a line integral. A simpler derivation would leave fixed and then observe that

It is doubtful that the Pythagoreans would have been able to do the above proof but they knew how to compute the area of a triangle and were familiar with figurate numbers and the gnomon, a segment added onto a geometrical figure. All of these ideas predate calculus and are an alternative for the integral.
The proportional relation between the changes and their sides is at best an approximation, so how can one justify its use? The answer is the approximation gets better for smaller changes since the arc of the circle which cuts off c more closely approaches the tangent to the circle. As for the sides and triangles, no matter how many segments they are divided into the sum of these segments is always the same. The Pythagoreans were trying to understand change and motion and this led them to realize that the number line was infinitely divisible. Could they have discovered the approximation for the changes in the sides? One only has to observe that the motion of the shadow of a sundial produces the hypotenuses of the triangles to derive the figure shown.

微分方程式證法
For a proof by the methods of rational trigonometry, see Pythagorean theorem proof (rational trigonometry).

Rational trigonometry
將畢氏定理做逆向推論後仍然可保留其正確性:
a. We need to prove that the angle between the a and b sides is a right angle. We construct another triangle with a right angle between sides of lengths a and b. By the Pythagorean theorem, it follows that the hypotenuse of this triangle also has length c. Since both triangles have the same side lengths a, b and c, they are congruent, and so they must have the same angles. Therefore, the angle between the side of lengths a and b in our original triangle is a right angle.
畢氏定理的逆定理是判斷三角形為鈍角、銳角或直角的一個簡單的方法，其中c為最長邊:

如果，則△ABC是直角三角形。
如果，則△ABC是銳角三角形。
如果，則△ABC是鈍角三角形。

應用及推廣
A Pythagorean triple consists of three positive integers a, b, and c, such that a. In other words, a Pythagorean triple represents the lengths of the sides of a right triangle where all three sides have integer lengths. Evidence from megalithic monuments on the British Isles shows that such triples were known before the discovery of writing. Such a triple is commonly written (a, b, c), some well-known examples are (3, 4, 5) and (5, 12, 13).

Pythagorean triples
One of the consequences of the Pythagorean theorem is that irrational numbers, such as the square root of two, can be constructed. A right triangle with legs both equal to one unit has hypotenuse length square root of two. The Pythagoreans proved that the square root of two is irrational, and this proof has come down to us even though it flew in the face of their cherished belief that everything was rational. According to the legend, Hippasus, who first proved the irrationality of the square root of two, was drowned at sea as a consequence.

無理數的存在
The distance formula in Cartesian coordinates is derived from the Pythagorean theorem. If (x₀, y₀) and (x₁, y₁) are points in the plane, then the distance between them, also called the Euclidean distance, is given by

More generally, in Euclidean n-space, the Euclidean distance between two points and , is defined, using the Pythagorean theorem, as:

在座標上的距離
The Pythagorean theorem was generalised by Euclid in his Elements:
If one erects similar figures (see Euclidean geometry) on the sides of a right triangle, then the sum of the areas of the two smaller ones equals the area of the larger one.
The Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines:

where θ is the angle between sides a and b.
When θ is 90 degrees, then cos(θ) = 0, so the formula reduces to the usual Pythagorean theorem.
Given two vectors v and w in a complex inner product space, the Pythagorean theorem takes the following form:

In particular, ||v + w||

泛論
畢氏定理是衍生自歐幾里得幾何的公設, 事實上, the Euclidean form of the Pythagorean theorem given above does not hold in non-Euclidean geometry. (It has been shown in fact to be equivalent to Euclid's Parallel (Fifth) Postulate.) For example, in spherical geometry, all three sides of the right triangle bounding an octant of the unit sphere have length equal to π / 2; this violates the Euclidean Pythagorean theorem because .
This means that in non-Euclidean geometry, the Pythagorean theorem must necessarily take a different form from the Euclidean theorem. There are two cases to consider — spherical geometry and hyperbolic plane geometry; in each case, as in the Euclidean case, the result follows from the appropriate law of cosines:
For any right triangle on a sphere of radius R, the Pythagorean theorem takes the form

By using the Maclaurin series for the cosine function, it can be shown that as the radius R approaches infinity, the spherical form of the Pythagorean theorem approaches the Euclidean form.
For any triangle in the hyperbolic plane (with Gaussian curvature −1), the Pythagorean theorem takes the form

where cosh is the hyperbolic cosine.
By using the Maclaurin series for this function, it can be shown that as a hyperbolic triangle becomes very small (i.e., as a, b, and c all approach zero), the hyperbolic form of the Pythagorean theorem approaches the Euclidean form.
In hyperbolic geometry, for a right triangle one can also write,

where is the angle of parallelism of the line segment AB that μ(AB) = a where μ is the multiplicative distance function (see Hilbert's Arithmetic of Ends).
In hyperbolic trigonometry, the sine of the angle of parallelism satisfies

Thus, the equation takes the form

where a, b and c are multiplicative distance of the sides of the right triangle (Hartshorne, 2000).

非歐幾何中的畢氏定理

In The Wizard of Oz, when the Scarecrow receives his diploma from the Wizard, he immediately exhibits his "knowledge" by reciting a mangled and incorrect version of the theorem: "The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side. Oh, joy, oh, rapture. I've got a brain!" The "knowledge" exhibited by the Scarecrow is incorrect. The accurate statement would have been "The sum of the squares of the legs of a right triangle is equal to the square of the remaining side."
In the Major-General's Song, "About binomial theorem I'm teeming with a lot o' news, With many cheerful facts about the square of the hypotenuse."

涉及畢氏定理的文化

直角三角形
勾股數
餘弦定理