By I. Martin Isaacs

I. Martin Isaacs', writer of the well known ALGEBRA: A GRADUATE path, has written a brand new textual content that may make readers have fun with the wonderful thing about geometry, specifically what it may possibly train approximately deductive reasoning and the character of mathematical proofs. the writer specializes in the 2 vintage topics of geometry: proof and proofs.

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22, the line segment that results is a diameter of the circle. Repeat this process to draw a second diameter and mark the point where the two diameters cross. This is clearly the center of the circle. Observe that given any line segment A B, there is a unique circle having A B as a diameter. This, of course, is the circle centered at the midpoint of A B and having radius � A B . 22, therefore, is that L C is a right angle in flA B C if and only if point C lies on the unique circle having side A B as a diameter.

On the right, we see another square of side a + b decomposed this time into one square and four right triangles. Again, each of the right triangles is congruent to the given one by SAS, and so the side length of the smaller square is exactly c. It follows that the area remaining when four copies of our triangle are removed from a square of side a + b is c 2 . We just saw, however, that this area is equal to a 2 + b 2 , and thus a 2 + b 2 == c2 , as required. There is one crucial detail we have omitted.

To see that the polygon formed by the n equally spaced points is regular, we must also establish that the n angles are all equal. Each of the n arcs is clearly equal in degrees to 360/ n degrees, and each angle of the polygon subtends an arc consisting of n - 2 of these small arcs. It follows by Theorem 1 . 1 6 that each of these angles is equal to � (n - 2) (360/n) == 1 80(n - 2)/n degrees, and thus the polygon is regular, as desired. Note that the sum of all the angles of our regular n-gon is n times 1 80(n - 2) / n.