About Pythagorean theorem

Pythagorean theorem in mathematics or Pythagorean Theureum is a relation between the three arms of the right triangle in the Euclidean geometry. This theorem is named after Greek mathematician Pythagoras, who are traditionally considered to be the inventor and proof of these theorems. However, the concept of the theorem was prevalent before its time. In China, this theorem is known as "Gauju Theurem" (勾股定理), which is applicable to 3, 4 and five arms triangle. In this sub-formula, the area of ​​the square shaped on the hypogeuse of a right triangle is equal to the sum of the area of ​​the two squares on the two sides of the triangle. If we take c the length of c and the other two lengths of a and b, then the theorem will be with the equation.

{\ displaystyle a ^ {2} + b ^ {2} = c ^ {2} ,} a ^ 2 + b ^ 2 = c ^ 2 ,
Or, in the case of standard c:

{\ displaystyle c = {\ sqrt {a ^ {2} + b ^ {2}}}. ,} c = \ sqrt {a ^ 2 + b ^ 2} ,
In this formula, a feature of the equilateral triangle is expressed with the help of a common formula, through which the length of the two sides of a triangle is known, then the length of the third arm can be determined. A common form of this formula is the law of cognition with which the length of the third arm of any triangle can be determined when the length of the remaining two arms and the angle between them is given. If the angle between the arms is angled, then it is possible to determine with the help of Pythagoras theorem.

Proof of using similar triangle

This proof is based on the proportion of which two duplicate triangles have been used.

Suppose ABC is a right-angled triangle, with the right angle C, shown in the figure. C curved perpendicular H intersects the arm, AB. The resulting new triangle ACH will be similar to the aforementioned ABC, as both of them have an angle angle and a angle A normal. This will equal the third angle, and for the same reason the CBH triangle is similar to ABC. Due to this similarity, two proportion ...

Will be

{\ displaystyle BC = a, AC = b, {\ mbox {and}} AB = c, ! BC = a, AC = b, \ mbox {and} AB = c, !
So

{\ displaystyle {\ frac {a} {c}} = {\ frac {hb} {a}} {\ mbox {and}} {\ frac {b} {c}} = {\ frac {AH} {b} }}. .} \ frac {a} {c} = \ frac {hb} {a} \ mbox {and} \ frac {b} {c} = \ frac {AH} {b}. ,
These are written in the following ways

{\ displaystyle a ^ {2} = c \ times HB {\ mbox {and}} b ^ {2} = c \ times AH. ,} a ^ 2 = c \ times HB \ mbox {and} b ^ 2 = c \ times AH. ,
It is found by adding two equations

{\ displaystyle a ^ {2} + b ^ {2} = c \ times HB + c \ times AH = c \ times (HB + AH) = c ^ {2}. , !} a ^ 2 + b ^ 2 = c \ times HB + c \ times AH = c \ times (HB + AH) = c ^ 2. , !
This is the Pythagorean theorem:

{\ displaystyle a ^ {2} + b ^ {2} = c ^ {2}. , !} a ^ 2 + b ^ 2 = c ^ 2. , !
Algebraic Evidence
Under the algebraic way the formula can be proved. The adjacent image has four right angle triangles in the four corners of the upper class, each of which is the area

{\ displaystyle {\ frac {1} {2}} AB.} \ frac {1} {2} AB

A square created by aligning four right angle triangles and a large square.
The corners of the A-side of the triangle and B are complementary to each other, so each angle in the middle blue area is a right angle. That is, the middle blue area is a square with the length of each arm C. C2 of the square As a result, the area of ​​the whole area is:

{\ displaystyle 4 \ left ({\ frac {1} {2}} AB \ right) + C ^ {2}.} 4 \ left (\ frac {1} {2} AB \ right) + C ^ 2
Meanwhile, the length of a arm of the upper class is A + B. Its area (A + B) 2 increases which increases to A2 + 2AB + B2. {\ displaystyle A ^ {2} + 2AB + B ^ {2} = 4 \ left ({\ frac {1} {2}} AB \ right) + C ^ {2}. , !} A ^ 2 + 2AB + B ^ 2 = 4 \ left (\ frac {1} {2} AB \ right) + C ^ 2. , !

(Distribution of the 4) {\ displaystyle A ^ {2} + 2AB + B ^ {2} = 2AB + C ^ {2} , !} A ^ 2 + 2AB + B ^ 2 = 2AB + C ^ 2 , !
(Subtract 2AB) {\ displaystyle A ^ {2} + B ^ {2} = C ^ {2} , !} A ^ 2 + B ^ 2 = C ^ 2 , !