Let us mark off two points on a straightedge so as to mark off a certain distanc

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Let us mark off two points on a straightedge so as to mark off a certain distance d. Archimedes showed how we can then trisect an arbitrary angle: For any angle, draw a circle beta of radius d centered at the vertex O of the angle. This circle cuts the sides of the angle at points A and B. Place the marked straightedge so that one mark gives a point C on line OA such that O is between C and A, the other mark gives a point D on circle beta, and the straightedge must simultaneously rest on the point B, so that B, C, and D are collinear. Prove that angle COD so constructed is one-third of angle AOB. (Hint: Use Euclidean theorems on exterior angles and isosceles triangles.)

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