Math 409/409G History of Mathematics

1. Math 409/409GHistory 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) 90o?

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 180o. 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 90o. 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 3.18. They are: C1.17: The non-right angles in a right triangle are each less than 90o. 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 3.18. 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 = 90o. (Def. ) • 1 < 90o. (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 assumptionthat 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