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Math 409/409G History of Mathematics Books III of the Elements Circles.

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Presentation on theme: "Math 409/409G History of Mathematics Books III of the Elements Circles."— Presentation transcript:

1 Math 409/409G History of Mathematics Books III of the Elements Circles

2 In Book III of the Elements, Euclid presented 37 propositions about circles. You are most likely familiar with many of these. For example: Proposition 3.31 states that an angle inscribed in a semicircle of a circle is a right angle.

3 But did you know that this proposition also says that if an angle is inscribed in a portion of a circle that is greater than (less than) a semicircle, then the angle is less than (greater than) 90 o ?

4 A proposition you may not be familiar with is Proposition 3.1 which states that it is possible to find (construct) the center of a given circle. Here’s how you do it.

5 Construct a segment joining two random points A and B of the circle. Construct the midpoint M of AB. Construct the perpendicular to AB at M and let it intersect the circle at C and D. Construct the midpoint O of CD. Euclid used an indirect proof to show that O is the center of the circle.

6 The proofs of most of the propositions in Book III use only the propositions from Book I. One such proposition is: Proposition 3.18: A tangent to a circle is perpendicular to the radius from the center to the point of tangency.

7 Before looking at the proof of this proposition, let’s review the significance of two of the Book I propositions used in the proof.

8 Proposition 1.17: The sum of the angles of a triangle is 180 o. The significance of this theorem is that when it is applied to a right triangle, it results in justifying that the non-right angles (  1 and  2) in the triangle must be less than 90 o. Today, this fact would be stated as a corollary to Proposition 1.17.

9 Proposition 1.19: In any triangle, the greater side is subtended by the greater angle. As you just saw, Proposition 1.17 shows that the greatest angle in a right triangle is the right angle. So as a consequence (corollary) of this proposition, we know that the hypotenuse of a right triangle is the greatest side of the triangle.

10 In modern terms, we now have two corollaries which will be used in the proof of Proposition They are: C1.17: The non-right angles in a right triangle are each less than 90 o. C1.19: The hypotenuse of a right triangle is greater than either leg of the triangle.

11 We are now ready to sketch the proof of Proposition Given: AB is tangent to circle O at point T. Prove: OT  AB.

12 By way of contradiction, assume that OT is not perpendicular to AB. Construct OC  AB. (P1.31)  2 = 90 o. (Def.  )  1 < 90 o. (C1.17) (C1.19)

13 But if D is the intersection of OC with the circle, then (Def. circle). So (CN 5).

14 But these last two statements are a contradiction. So the assumption that OT is not perpendicular to AB cannot be true. Thus it must be true that OT is indeed perpendicular to AB.

15 This proves that a tangent to a circle is perpendicular to the radius from the center of the circle to the point of tangency. This proof also shows you that a “proof by contradiction” doesn’t always have to contradict the hypothesis of the theorem.

16 This ends the lesson on Books III of the Elements Circles


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