We will get 4 right-angled triangle, hypotenuse of each of them is ‘a’: remaining sides of each of them are band c. Remaining part of the figure is the
square EFGH, each of whose side is a, so area of the square EFGH is a2.
Now, we are sure that square WXYZ = square EFGH + 4 ∆ GYF
or, (b + c)2 = a2 + 4 ∙ 1/2 b ∙ c
or, b2 + c2 + 2bc = a2 + 2bc
or, b2 + c2 = a2
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