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Several Proof Methods of Pythagorean Theorem
1, make eight congruent right-angled triangles, let their two right-angled sides be A and B respectively, and the hypotenuse be C, and then make three squares with sides A, B and C respectively to make the two squares shown in the above figure.

As can be seen from the figure below, the sides of these two squares are both a+b, so the areas are equal. That is, the square of A plus the square of B, plus 4 times half ab equals the square of C, plus 4 times half ab equals the square of C. ..

2. Make four congruent right-angled triangles, where A and B are right-angled sides and C is hypotenuse, so the area of each right-angled triangle is equal to half of ab. Put these four right triangles into the shape shown in the figure, so that A, E and B are on a straight line, B, F and C are on a straight line, and C, G and D are on a straight line.

∫rtδHAE≌rtδEBF,

∴ ∠AHE = ∠BEF。

∫∠AEH+∠AHE = 90? ,

∴ ∠AEH + ∠BEF = 90? .

∴ ∠HEF = 180? ―90? = 90? .

∴ The side length of quadrilateral EFGH is C.

Square. Its area is equal to c2.

∫rtδGDH≌rtδHAE,

∴ ∠HGD = ∠EHA。

∫∠HGD+∠GHD = 90? ,

∴ ∠EHA + ∠GHD = 90? .

∫∠GHE = 90 again? ,

∴ ∠DHA = 90? + 90? = 180? .

∴ ABCD is a square with a side length of A+B, and its area is equal to the square of A+B.

∴a plus B squared equals 4 times 1/2 ab plus C squared.

∴ The square of A plus the square of B equals the square of C.

3. take a and b as right angles (B >;; A), with C as the hypotenuse, make four congruent right-angled triangles, so the area of each right-angled triangle is equal to half ab. Combine these four right triangles into the shape as shown in the figure.

∫rtδDAH≌rtδABE,

∴ ∠HDA = ∠EAB。

∵ ∠HAD + ∠HAD = 90? ,

∴ ∠EAB + ∠HAD = 90? ,

ABCD is a square with a side length of c and an area equal to c2.

∫EF = FG = GH = HE = b―a,

∠HEF = 90? .

∴ EFGH is a square with a side length of B-A, and its area is equal to the square of B minus A.

∴ 4 times half ab+, and the square of B minus A equals the square of C.

∴ A 2+B 2 = C 2 (indicating that A 2 is the square of A).

4. Make two congruent right-angled triangles, A and B are right-angled sides and C is hypotenuse, so the area of each right-angled triangle is equal to half of ab. Put these two right triangles into the shape as shown in the figure, so that A, E and B are in a straight line.

∫rtδEAD≌rtδCBE,

∴∠ Ade = ∠BEC.

∫∠AED+∠ADE = 90? ,

∴ ∠AED + ∠BEC = 90? .

∴ ∠DEC = 180? ―90? = 90? .

∴δdec is an isosceles right triangle,

Its area is equal to half of c 2.

∫∠DAE = 90 again? ,∠EBC = 90? ,

∨ BC ∴.

∴ ABCD is a right-angled trapezoid with an area equal to 1/2 (a+b) 2.

∴ 1/2(a+b)^2=2x 1/2ab+ 1/2c^2.。

∴a^2+b^2=c^2.

5. Make four congruent right-angled triangles, set their two right-angled sides as A and B, and the hypotenuse as 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

a^2+b^2=S+2 x 1/2xab

C 2 = s+2x1/2x company

∴ a^2+b^2=c^2.

References:

Baidu encyclopedia-Pythagorean theorem