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Thread: Similar Triangles: Error in course or understanding?

  1. #1
    Newbie B9766's Avatar
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    Similar Triangles: Error in course or understanding?

    I'm have trouble with a particular problem in a Geometry course from The Great Courses. I've found other errors in the course and believe this is another but it's so egregious that I wanted to check to make sure I'm not just missing something. Please take a look at the problem below:

    Similar Triangles: Error in course or understanding?-problem-10.jpg

    Question 10.i asks, Use similar triangles to show that

    \frac{y}{y+x} = \frac{s}{s+t}

    Shouldn't that be:

    \frac{x}{y+x} = \frac{s}{s+t}

    Your help will be appreciated.
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  2. #2
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    Re: Similar Triangles: Error in course or understanding?

    Quote Originally Posted by B9766 View Post
    Click image for larger version. 

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    Question 10.i asks, Use similar triangles to show that
    \frac{y}{y+x} = \frac{s}{s+t}
    Shouldn't that be:
    \frac{x}{y+x} = \frac{s}{s+t}
    Actually it should be \dfrac{y}{y+x} = \dfrac{t}{s+t}
    See that the 'sub-triangle' is similar to the 'super-triangle'.
    Thanks from B9766
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  3. #3
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    Re: Similar Triangles: Error in course or understanding?

    I would feel better about \frac{y}{y+ x}= \frac{t}{s+ t} since y and t are sides of a triangle while x ad s are not so this follows directly from "similar triangles".

    However, from \frac{y}{y+ x}= \frac{t}{s+ t}, we get 1- \frac{y}{y+ x}= 1- \frac{t}{s+ t} and then \frac{y+ x- y}{y+ x}= \frac{s+ t- t}{s+ t} so \frac{x}{x+y}= \frac{s}{s+ t} is also true.
    Last edited by HallsofIvy; Feb 23rd 2017 at 12:17 PM.
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    Newbie B9766's Avatar
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    Re: Similar Triangles: Error in course or understanding?

    Thank you! I agree. I was so lost in segments and sub-segments I lost sight of the triangles.
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