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Unformatted text preview: H F C F A F B B C B A C A Problem 3) To Prove: M C Q A Q C is a right angle. Proof: Let J be the point where the altitude BF B and the segment Q A Q C intersect. Since BF B is an altitude, BF B AC , and so BF B C is a right angle. Also, from the proof in Problem 2, Q A Q C AC . Therefore, by the a corollary of the Converse of the Alternate Interior Angles Tbeorem, BF B C BJQ C . So, BJQ C is a right angle. From the proof in Problem 1, M C Q A BH . Therefore, by the a corollary of the Converse of the Alternate Interior Angles Tbeorem, M C Q A Q C BJQ C . Therefore, M C Q A Q C is a right angle. QED HW 7A, Part II Solutions (Continued): J M B M C M A Q C Q B Q A H F C F A F B B A C...
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- Spring '10