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In general, b can be any value. For our present proof, we choose b < a.

Take a square ABCD of side a. Its area is a². Mark K and F on AB and AC respectively at distance b. Draw lines perpendicular to AB and BC from K and F respectively to meet at E.  KBFE is a square with side equal to b. Its area is b².

Now mark all sides and segments.  FC = a-b = GE = EH = HD.  So DHEG is a square with side (a-b). Its area is (a-b)².

There are two rectangles HAKE and GEFC. Each of them has sides (a-b) and b. Their areas are b(a-b) = ba - b².

Now,   Area of ABCD = Area of HAKE + GEFC + KBFE + DHEG

                            a² = ba - b² + ba -b² + b² + (a-b)²= 2ab - b² + (a-b)²

using standard algebra we get,   (a-b)² = a² + b² - 2 a b

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