hw-8-wksheet-8a-soln - B , H , and F C . [ Part II ] A H F...

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[ Part III ] By definition of "altitude", B F C H and B F A H are right angles. B F C H and B F A H are two right triangles with common hypotenuse BH . By the Common Hypotenuse Theorem (Thm (NIB) 4.7), the circle C( diam = BH ) contains the four points B , F C , H , and F A . The two angles F A H B and B F C F A both are inscribed in C( diam = BH ) and they intercept the same arc BF A . By a corollary of the Inscribed Angle Theorem, F A H B 2245 B F C F A . [ Part IV ] A F C F B 2245 A H F B from Part I . A H F B 2245 F A H B from Part II . F A H B 2245 B F C F A from Part III . A F C F B 2245 B F C F A by transitivity. [ Q E D for (2) ] Proof: Proof of (2): [ Part I ] (See figure below.) By definition of "altitude", A F C H and A F B H are right angles. A F C H and A F B H are two right triangles with common hypotenuse AH . By the Common Hypotenuse Theorem (Thm (NIB) 4.7), the circle C( diam = AH ) contains the four points A , F
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Unformatted text preview: B , H , and F C . [ Part II ] A H F B and F A H B are vertical angles, so A H F B 2245 F A H B , by the Vertical Angles Theorem. The two angles A F C F B and A H F B both are inscribed in C( diam = AH ) and they intercept the same arc AF B . By a corollary of the Inscribed Angle Theorem, A F C F B 2245 A H F B . Worksheet 8A Solution: Proof of statement (2) in the Problem Problem: It is given that ABC is an acute triagle where F A , F B , and F C are the feet of the altitudes from A, B, and C, resp. It is also given that these three altitudes are concurrent at a point H in the interior of the triangle. To Prove: (1) B F A F C 2245 C F A F B (2) A F C F B 2245 B F C F A (3) C F B F A 2245 A F B F C H F C F B F A F B F C H F A C A B C A B...
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This note was uploaded on 11/02/2010 for the course MATH modern geo taught by Professor Shirley during the Spring '10 term at University of Texas at Austin.

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