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Thread: Prove - medial triangle, perpendicular bisector, altitude

  1. #1
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    Prove - medial triangle, perpendicular bisector, altitude

    The problem is to:

    (1) Prove that $\displaystyle \Delta$ABC is the medial triangle for $\displaystyle \Delta$A'B'C'.

    (2) Prove that the perpendicular bisector of a side of $\displaystyle \Delta$A'B'C' is an altitude for $\displaystyle \Delta$ABC.

    (3) Prove that the three altitudes of any triangle are concurrent.

    The given information is:

    Given a triangle $\displaystyle \Delta$ABC, draw a line through each vertex that is parallel to the opposite side. The three lines intersect to form a new triangle $\displaystyle \Delta$A'B'C'. The diagram can not be used as the proof.

    How do you write this proof out without using the diagram?
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  2. #2
    MHF Contributor red_dog's Avatar
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    a) $\displaystyle C'ACB$ is a parallelogram. Then $\displaystyle C'A=BC$

    $\displaystyle AB'CB$ is also a parallelogram, then $\displaystyle AB'=BC$.

    Then $\displaystyle C'A=AB'$ and $\displaystyle A$ is the midpoint of $\displaystyle B'C'$.

    Do the same for the others sides of the triangle $\displaystyle A'B'C'$

    b) Let $\displaystyle AM\perp BC$. But $\displaystyle BC\parallel B'C'\Rightarrow AM\perp B'C'$. Then $\displaystyle AM$ is the perpendicular bisector of $\displaystyle B'C'$.

    c) The perpendicular bisectors of the sides of the triangle $\displaystyle A'B'C'$ are concurrent, so the altitudes of the triangle $\displaystyle ABC$ are concurrent.
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