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