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Thread: Special Segments in Triangles

  1. #1
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    Special Segments in Triangles

    Special Segments in Triangles-untitled-drawing.jpg
    This was on a special segments sheet, I know the answer is 9, but I don't know how to prove it.
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  2. #2
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    Re: Special Segments in Triangles

    Call the point where the vertical line intersects BC point O. Assuming the vertical line is perpendicular to BC (is it?), then angle BOA = COA = 90 degrees. Then you have triangles ABO and ACO are equivalent from side-angle-side. Hence AB = AC.
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    Re: Special Segments in Triangles

    Quote Originally Posted by kamnarra View Post
    Click image for larger version. 

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    This was on a special segments sheet, I know the answer is 9, but I don't know how to prove it.
    I see no reason that it should equal 9. I see no reason to assume there are any right angle(s) in the figure.
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    Re: Special Segments in Triangles

    Sorry, there was a right angle. I forgot to draw it.
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    Re: Special Segments in Triangles

    Quote Originally Posted by kamnarra View Post
    Sorry, there was a right angle. I forgot to draw it.
    Well then, next time please review before posting making sure all relevant information is given.
    The only triangle in which an altitude is a bisector of its base is an isosceles triangle.
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