Mamikon’s proof of the Pythagorean Theorem

Consider two concentric circles of radii a and c, with a < c. Let b be the semi-chord of the larger circle tangent to the smaller circle. Note that a, b, and c can be arranged so as to form a right triangle with legs a and b, and hypotenuse c. Note also that any given right triangle can be retro-fitted to this situation. Then, the the area of the annulus (the region between the two circles) can be shown, without use of the Pythagorean Theorem, to be (pi)(c^2 – a^2). However, as b makes one revolution, it sweeps out the same area which, by Mamikon’s Theorem, is equal to the corresponding tangent cluster, which has area (pi)b^2. Therefore, (pi)(c^2 – a^2) = (pi)b^2.
Therefore, c^2 = a^2 + b^2. Q.E.D. Thus, Mamikon has the distinction of coming up with A NEW PROOF OF THE PYTHAGOREAN THEOREM.
keywords: Mathematics, Geometry, Visual Calculus