hw-8-4.7-no-9-soln - Solution for HW #8A, Part II,...

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Similarly, F C F B H 2245 H F B F A and F B F C H 2245 H F C F A . Therefore, segment F A H is the angle bisector of F C F A F B , segment F B H is the angle bisector of F C F B F A , and segment F C H is the angle bisector of F B F C F A . Thus, H is the point of concurrence of the angle bisectors of the interior angles of F A F B F C . Thus, H is the Incenter of F A F B F C . Q E D Method 1: By the Common Hypotenuse Theorem applied twice, circle c 1 = C(diam = AC ) contains A, C, F A , and F C , and circle c 2 = C(diam = HC ) contains A, H, F A , and F B . See the figure. Note that F C F A H = F C F A A , so F C F A H 2245 F C F A A . F C F A A and F C C A are both inscribed in circle c 1 and they intercept the same arc (F C A ) , so by a corollary to the Inscribed Angle theorem , F C F A A 2245 F C C A . Note that F C C A = H C F B , so F C C A 2245 H C F B , Now, H C F B and H F A F B are both inscribed in circle c
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