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利用如下拼图方法可以证明勾股定理:由4个全等的小长方形拼接成如下图的一个大正方形ABCD,并假设每个小长方形长=a,宽=b,对角线=c,则大正方形ABCD的边长为(a+b),且中空部分IJKL为边长为(a-b)的小正方形.连结对角线EH、HG、GF、FE、IK,并如图设出∠1、∠2、∠3、∠4、∠5、∠6.令Ⅰ=△IKL,Ⅱ=△HEL.c∵△HAE≌△EBF,∴∠1=∠5,∠2=∠6.而∠3+∠4=∠1+∠6=∠5+∠6=90°同理可证明∠EHG=∠HGF=∠GFE=90°,故四边形EFGH为正方形,且边长为c.又∵SI+4SⅡ=SABCD/2,而SEFGH-SI=SABCD/2,∴2SI+4SⅡ=SEFGH.而SI=(a-b)2/2,SⅡ=ab/2,SEFGH=
The Pythagorean Theorem can be proved using the following puzzle method: a large square ABCD of four congruent small rectangles is assembled, and assuming that each small rectangle has length = a, width = b and diagonal = c, The side length of the square ABCD is (a + b), and the hollow part IJKL is a small square with side length (ab). Connect the diagonal lines EH, HG, GF, FE, IK and set ∠1, ∠ 2, ∠3, ∠4, ∠5, ∠6. Let I = △ IKL, Ⅱ = △ HEL.c∵ △ HAE≌ △ EBF, ∴∠1 = ∠5, ∠2 = ∠6 and ∠3 + ∠ 4 = ∠ 1 + ∠ 6 = ∠ 5 + ∠ 6 = 90 ° Similarly we can prove that ∠ EHG = ∠ HGF = ∠ GFE = 90 °, so the quadrilateral EFGH is square and the side length of c. = SABCD / 2, and SEFGH-SI = SABCD / 2, ∴2SI + 4SII = SEFGH, while SI = (ab) 2/2, SII = ab / 2, SEFGH =