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The Pythagorean Theorem B c a C A b Figure 1: Given a right triangle △ABC with vertices A, B, C opposite sides of length a, b, c, respectively, and 6 C = 90◦ a right angle, the Pythagorean Theorem says that c2 = a2 + b2 . Here is a simple proof. Proof of the Pythagorean Theorem. Without loss of generality we may assume that the sides are labeled so that a ≤ b. c2 is the area of a square with sides c, so let us begin by drawing a square with sides c. Draw a line through each corner of the square as in fig. (2.i) below, so that the angle between the line and the corresponding side of the square is congruent to 6 A. Now 6 C = 90◦ and 180◦ = 6 A + 6 B + 6 C so 90◦ = 6 A + 6 B. Therefore the lines divide each corner of the square into two angles, one congruent to 6 A and the other congruent to 6 B. (See fig. (2.ii)). c A A c A a c B b c b−a B c A B b−a b−a a A A a b−a b b c b A a c B c i) A ii) Figure 2: It follows that each of the four triangles in in fig. (2.ii) has a side of length c joining an angle congruent to 6 A with an angle congruent to 6 B. By the 1 “angle-side-angle” theorem in elementary geometry it follows that the four triangles in fig. (2.ii) are all congruent to △ABC. Therefore the angles of the quadrilateral in the center of fig. (2.ii) are all right angles. Clearly the sides of the quadrilateral all have length b − a so the quadrilateral is a small square with area (b − a)2 . Again referring to fig. (2.ii), we see that (area of big square) = 4(area of triangle) + (area of small square) hence c2 = 4 ab + (b − a)2 . 2 Multiplying out we have c2 = 2ab + (b2 − 2ab + a2 ) hence c2 = b2 + a2 . This completes the proof. 2