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Unformatted text preview: ACB is a right angle (see Figure 1.1.3(a)). In other words, the triangle ABC is a right triangle. A B C O (a) A B C O (b) Figure 1.1.3 Thales Theorem: ACB = 90 To prove this, let O be the center of the circle and draw the line segment OC , as in Figure 1.1.3(b). Let = BAC and = ABC . Since AB is a diameter of the circle, OA and OC have the same length (namely, the circles radius). This means that OAC is an isosceles triangle, and so OCA = OAC = . Likewise, OBC is an isosceles triangle and OCB = OBC = . So we see that ACB = + . And since the angles of ABC must add up to 180 , we see that 180 = + ( + ) + = 2( + ), so + = 90 . Thus, ACB = 90 . QED...
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This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.
 Fall '11
 Dr.Cheun
 Calculus, Angles

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