The Pythagorean theorem is proved by the method of complementarity of entry and exit

The side length of the square ABCD is A, the point B is on the AG, the side length of the square EFGB is B, the point C is on the EB, the side length of the square EHIA is C, the point H is on the FG, let the IJ AG be J, the Hi will be on K, and the AE will be Cd on L;

ea=eh=a,eb=ef=b, eba= efh=90°, rt efh rt eba, 1= 2, fh=ba=a , rt efh, right-angled edge fh=a, right-angled edge ef=b, hypotenuse eh=c , 2= 3= 4=90°- eab, 1= 2, 1= 3,eh=ai=a, efh= aji=90°, rt efh rt aji,ji=fh=a , 5= 3=90°- aij, 3= 4 , 4= 5, da=ji=a, adl= ijk=90°, rt adl rt ijk, 6= 1=90°- ehf, 1= 2 , 2= 6, ec=hb=b-a, lce= kgh=90°

rt△lce≌rt△kgh ；

To sum up: square ABCD area + square EFGB area.

square ehia area;

i.e.: a +b = c ;

In a right triangle, the sum of the squares of the two right-angled sides is equal to the square of the hypotenuse.

Let's draw a right-angled triangle first, and then add a square to the side of the triangle next to the shortest right-angled side, which is indicated in red for clarity. Add another square below another right-angled side, indicated in blue. Next, draw a square at the length of the hypotenuse, as shown in Figure 5(b).

We are going to prove that the sum of the areas of the red and blue squares is exactly equal to the area of the square drawn on the hypotenuse.

Note in Figure 5(b) that when the hypotenuse square is added, the red and blue parts of the square are partially outside the range of the hypotenuse square. Now I'm going to show the out-of-range parts in yellow, purple, and green, respectively. At the same time, there are some parts of the hypotenuse square that have not been filled in with color.

Now, according to the method in Figure 5(c), move the out-of-range triangle into the uncolored area. We found that the out-of-range part was just filling in the uncolored area! From this, we can find that the sum of the areas of the red and blue parts in Figure 5(a) must be equal to the area of the hypotenuse square in Figure 5(c).

From this, we have confirmed the Pythagorean theorem.

Look at the picture above. The triangle is a right-angled triangle, the square with hook a as the side is the zhu square, and the square with strand b as the side is the green square. To make up for the deficiency with surplus, Zhu Fang and Qing Fang are merged into Xuan Fang.

According to its area relationship, there is a + b = cSince Zhu Fang and Qing Fang each had a part in the Xuanfang, that part did not move.

The square with the hook as the edge is the Zhu square, and the square with the strand as the edge is the green square. In order to win and make up for the void, as long as the i of Zhu Fang (A2) in the figure is moved to I, the II of Qing Fang is moved to II, and III is moved to III, then a square (C2) with the string as the side can be put together to obtain A2+B2=C2

Liu Hui used the "entry and exit complement method", that is, the cut-and-paste proof method, he cut out some areas on the square with the Pythagorean side (out) and moved it to the blank area of the square with the string as the edge (in), and the result was just filled, and the problem was solved completely with the ** method. The following ** is Liu Hui's "Qingzhu entry and exit map."

The arrows indicate the translation of the triangle.

The idea of proof is as follows: prove that the area of the square AEHI is equal to the area of the square EBGF plus the area of the square ABCD.

Proof process: 1. Prove the congruence of the top two small right-angled triangles.

2. Prove that the two small right-angled triangles in the lower right corner are congruent.

3. Prove that the great right triangle on the far right and the great right triangle at the bottom are congruent.

4. It is enough to prove that the area is equal.

Just move it, it's in the book.

Amazing stuff. I don't understand.

The method of complementing the entry and exit is used to prove the theorem of the vertical hand of the Pythagorean vertical hand.

a.Zu Chongzhi.

b.Zhang heng. c.Liu Hui.

d.Zhen Luan. Correct Answer: Residual Closure C

If it is proved that 1 and 2 are equal, then the method of proving that b=2a proves that the area is equal seems to be the only way to prove congruence.

So when celery b=2a, you can use AAS to prove 1 2, s1=s2<>

The side length of the square ABCD is A, the point B is on the AG, the side length of the square EFGB is B, the point C is on the EB, the side length of the square EHIA is C, the point H is on the FG, let the IJ AG be J, the Hi will be on K, and the AE will be Cd on L;

ea=eh=a,eb=ef=b, eba= efh=90°, rt efh rt eba, 1= 2, fh=ba=a , rt efh, dan wang right-angled edge fh=a, right-angled edge ef=b, hypotenuse eh=c , 2= 3= 4=90°- eab, 1= 2, 1= 3, eh=ai=a, efh= aji=90°, rt efh rt aji,ji=fh=a , 5= 3=90°- aij, 3= messy4 , 4= 5, da=ji=a, adl= ijk=90°, rt adl rt ijk, 6= 1=90°- ehf, 1= 2 , 2= 6, ec=hb=b-a, lce= kgh=90°

rt△lce≌rt△kgh ；

To sum up: square ABCD area + square EFGB area.

square ehia area;

i.e.: a +b = c ;

In a right-angled triangle, the sum of the squares of the two right-angled sides is equal to the square of the hypotenuse.

As shown in the figure: the side length of the square ABCD is A, the point B is on the AG, the side length of the square EFGB is B, the point C is on the EB, the side length of the square EHIA is C, the point H is on the FG, let the IJ AG be J, the Hi is on K, and the AE is Cd on L; ea=eh=a,eb=ef=b, eba= efh=90°, rt efh rt eba, 1= 2, fh=ba=a , rt efh, straight entangular edge fh=a, right angle edge ef=b, hypotenuse eh=c , 2= 3= 4=90°- eab, 1= 2, 1= 3, eh=ai=a, efh= aji=90°, rt efh rt aji,ji=fh=a , 5= 3=90°- aij,tuansou 3= 4 , 4= 5,da=ji=a, adl= ijk=90°, rt adl rt ijk, 6= 1=90°- ehf, 1= 2 , 2= 6,ec=hb=b-a, lce= kgh=90° rt lce rt kgh ; To sum up: square ABCD area + square EFGB area = square EHIA area; Namely:

a²+b²=c² ；In a right triangle, the sum of the squares of the two right-angled sides is equal to the square of the hypotenuse.