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Math Help - proportions

  1. #1
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    proportions

    Hello, I have a hard time figuring out where to start for this problem:

    If  \frac{a}{b} = \frac{c}{d} prove that

     \frac{ a^2c + ac^2}{b^2d + bd^2} = \frac{(a+c)^3}{(b+d)^3}
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  2. #2
    Junior Member roy_zhang's Avatar
    Joined
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    Quote Originally Posted by Percent09 View Post
    Hello, I have a hard time figuring out where to start for this problem:

    If  \frac{a}{b} = \frac{c}{d} prove that

     \frac{ a^2c + ac^2}{b^2d + bd^2} = \frac{(a+c)^3}{(b+d)^3}
    From given, we know b\ne 0,\;d\ne 0. Now we consider 2 cases for this proof:

    (1) If a=c=0, the prove is trivial.

    (2) Now let's consider the case a\ne 0 and c\ne 0. By a simply manipulation, we found that showing  \frac{ a^2c + ac^2}{b^2d + bd^2} = \frac{(a+c)^3}{(b+d)^3} is equivalent to showing  \frac{ac}{bd} = \frac{(a+c)^2}{(b+d)^2} . Now let's start from the only given identity and see what we can get.

      \frac{a}{b} = \frac{c}{d}\Rightarrow  \frac{a}{c} = \frac{b}{d}\Rightarrow \frac{a+c}{c}=\frac{b+d}{d}\Rightarrow \frac{a+c}{b+d}=\frac{c}{d}\;\;\dagger

    Also realize that \frac{a}{c} = \frac{b}{d}\Rightarrow\frac{c}{d}=\frac{a}{b}\;\;\  ddagger

    Combine \dagger and \ddagger, we have
     \frac{a+c}{b+d}=\frac{c}{d} and  \frac{a+c}{b+d}=\frac{a}{b}, multiply these two identities gives us  \frac{ac}{bd} = \frac{(a+c)^2}{(b+d)^2} .

    Roy
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