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Math Help - URGENT Geometry - triangle (angle bisector, median, altitude)

  1. #1
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    URGENT Geometry - triangle (angle bisector, median, altitude)

    Prove that a triangle in which the median and the altitude is symmetric with respect to the angle bisector from the same vertex must have a right angle in this vertex. (Please take a look at the image attached.)

    I can prove that if the triangle is right-angled, then the symmetry holds but I can't prove that if it isn't right-angled then they aren't symmetric.
    Attached Thumbnails Attached Thumbnails URGENT Geometry - triangle (angle bisector, median, altitude)-symmetry.png  
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    Senior Member vincisonfire's Avatar
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    AD = DB = DC = radius
    DB = DC gives us CDB = 180 - 2 * BCD
    AD = DB gives us ADC = 180 - 2 * ACD
    CDB + ADC = 180
    thus
    90 = ACD + BCD
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  3. #3
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    Yeah, I know that but what if \overline{DC}\ne r. Sorry for the misleading figure - C is not on the circle (if it were then the triangle would be right-angled - Thales' theorem).
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    Senior Member vincisonfire's Avatar
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    There is always a circle passing by three points. If it is not this one it's gonna be the other one?
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  5. #5
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    Yes but \overline{AD}\ne \overline{BD}. I mean the circumcenter won't be D, the midpoint of \overline{AB}.
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  6. #6
    Senior Member vincisonfire's Avatar
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    Yeah, if I get something I tell you.
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  7. #7
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    Thank you very much.
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