hw-7-7a-ii-solns - H F C F A F B B C B A C A Problem 3) To...

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Problem 2) To Prove: M C M A Q A Q C . Proof: M C is the midpoint of AB and M A is the midpoint of BC . Thus, by the "Midpoint Connection Theorem" (Theorem 4.2.15) applied to ABC, M C M A AC . Furthermore, Q A is the midpoint of AH and Q C is the midpoint of CH . Thus, by the "Midpoint Connection Theorem" (Theorem 4.2.15) applied to ACH, Q A Q C AC . Therefore, M C M A Q A Q C by Theorem 4.2.9. QED Furthermore, M A is the midpoint of BC and Q C is the midpoint of CH . Thus, by the "Midpoint Connection Theorem" (Theorem 4.2.15) applied to BCH, M A Q C BH . Therefore, M A Q C M C Q A by Theorem 4.2.9. QED Problem 1) To Prove: M A Q C M C Q A . Proof: M C is the midpoint of AB and Q A is the midpoint of AH . Thus, by the "Midpoint Connection Theorem" (Theorem 4.2.15) applied to ABH, M C Q A BH . HW 7A, Part II Solutions: Suppose ABC is an acute triangle. We adopt the F A -- M A -- Q A notation here. M B M C M A Q C Q B Q A H F C F B M B M C M A Q C Q B Q A
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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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hw-7-7a-ii-solns - H F C F A F B B C B A C A Problem 3) To...

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