Proofs of the Pythagorean Theorem
Proof 1. (Gardner 154)
Begin with the figure on the left below. Arrange four duplicated right triangles in this pattern. Now by rearranging the four triangles inside the largest square, the figure on the right can be reached. The two white squares are each squares on the legs of the triangle. Since their area is that of the large square minus the four triangles, we know it must equal the area of the white square on the left. Therefore, the square on the hypotenuse (right picture) has an area equal to the area of the sum of the squares on the legs (right picture).
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Proof 2. James A. Garfield's Proof. (Gardner 155)
Garfield stated "we think it something on which the members of both houses can unite without distinction of the party." This referred to his proof which uses the diagram below.
The basic right triangle is shaded. On the hypotenuse the right isosceles triangle CBE is drawn. Extend AC, then from a point E a perpendicular is drawn to the extended line, meeting at D. The shaded triangle is congruent with triangle DCE, therefore AB=DC and AC=DE.
From here, the area of trapezoid ABED is the product of the base and half the sum of its sides, x and y.
The area of the trapezoid is also the sum of the areas of the three triangles. These areas are z2/2, xy/2 and xy/2. So, another way of writing the area is z2 /2 + 2(xy)/2. By setting these two expressions equal to one another Garfield got x2+y2 = z2 . This was his proof of the Pythagorean theorem.