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hw-8-wksheet-8a-soln - B H and F C Part II ∠ A H F B and...

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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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