AB2+AC2=BC2

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Make 4 copies of the right triangle ABC. We can put two triangles together so that their hypotenuses are touching to form a rectangle. So now we have two rectangles. The area of each rectangle is AB.AC, that is, the product of the lengths of the legs. So the area of each triangle is half the area of the rectangle=½AB.AC. The total area of the 4 triangles is 2AB.AC.

Each triangle has a right angle at A. We are going to make two squares where the corners will be formed by the four As of the triangles. Lay the triangles so that the short leg of one triangle and the long leg of the next triangle form the side of a square. What you will have is an empty square formed by the hypotenuses of the four triangles inside a larger square formed by the combined lengths of the short and long legs of the triangles.

The side of the big square will have length AB+AC, and the inside square will have side length BC.

So the area of the smaller square is BC² and the area of the larger square is (AB+AC)². The total area of the triangles is 2AB.AC. The sum of the areas of the triangles and the area of the smaller square is equal to the area of the larger square, so:

BC²+2AB.BC=(AB+AC)²=AB²+2AB.AC+AC².

We can subtract 2AB.BC from each side and we are left with:

BC²=AB²+AC² or AB²+AC²=BC²

by Top Rated User (1.1m points)

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