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Math Help - 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 \DeltaABC is the medial triangle for \DeltaA'B'C'.

    (2) Prove that the perpendicular bisector of a side of \DeltaA'B'C' is an altitude for \DeltaABC.

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

    The given information is:

    Given a triangle \DeltaABC, draw a line through each vertex that is parallel to the opposite side. The three lines intersect to form a new triangle \DeltaA'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) C'ACB is a parallelogram. Then C'A=BC

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

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

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

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

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