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  1. #1
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    triangle problem

    BF=CF
    Angle DFC= Angle AFD
    Angle EFB= Angle AFE
    Prove that DE is Parallel to CB

    please help
    thanks in advance
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  2. #2
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    It is not true.

    If the line (which looks like altitute) is not perpendicular as it appears but rather slanated it fails.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    It is not true.

    If the line (which looks like altitute) is not perpendicular as it appears but rather slanated it fails.
    I'm glad someone else noted that. I had thought it to be true but couldn't prove that I wasn't missing something.

    -Dan
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  4. #4
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    10x
    Last edited by balmora55; March 17th 2007 at 11:01 PM.
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  5. #5
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    Quote Originally Posted by balmora55 View Post
    sorry to bother you but i have two problems
    1. explain what you said in greater detail please
    2. i don't know what altitute means
    (i'm from israel so i don't know math words in english)
    Heir.
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  6. #6
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    Quote Originally Posted by balmora55 View Post
    BF=CF
    Angle DFC= Angle AFD
    Angle EFB= Angle AFE
    Prove that DE is Parallel to CB

    ...
    Hello,

    I've attached a drawing of a triangle with the given conditions.

    (This sign < means angle)

    <(EFD) = 90 (it is the half of 180 and therefore constant)

    The intersection of AF ∩ DE = P

    Now reflect P over FD to P'. Then PP' is parallel to FE because PP' is perpendicular to FD and FD is perpendicular to FE. P' lies on CB because the triangle PFP' is an isosceles triangle. Therefore P'F must be parallel to PE because PEFP' is a parallelogramm.

    Do the same procedure with P, P'', F and D.

    You get DE is parallel to P'P'' and therefore DE is parallel to BC.

    EB
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