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Unformatted text preview: b a c A B C O D r E F Figure 2.5.6 Inscribed circle for ABC Let r be the radius of the inscribed circle, and let D , E , and F be the points on AB , BC , and AC , respectively, at which the circle is tangent. Then OD AB , OE BC , and OF AC . Thus, OAD and OAF are equivalent triangles, since they are right triangles with the same hypotenuse OA and with corresponding legs OD and OF of the same length r . Hence, OAD = OAF , which means that OA bisects the angle A . Similarly, OB bisects B and OC bisects C . We have thus shown: For any triangle, the center of its inscribed circle is the intersection of the bisectors of the angles....
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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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