Proof right angle triangle of Pythagoras?

Answer 1

Consider a square with a slightly smaller square just inside, the smaller square being rotated so that the corners just touch the sides of the larger square. The four spaces between the two squares are four exactly equal right angled triangles. If we call the sides of these triangles h=hypotenuse, L=long side and s=short side then we see that the area of the small square is h squared and the area of the large square is (s +L) squared =s squared + L squared +2s x L. The area of the four triangle is (4 x s x L)/2

By simple arithmetic:- the area of the small square (h squared) = The area of the large square minus the area of the four triangles, h squared = s squared +L squared + 2s x L - (4/2 x s x L). The last two terms cancel each other out leaving h squared = s squared + L squared.