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Thread: proportions

  1. #1
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    proportions

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

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

    $\displaystyle \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
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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 $\displaystyle \frac{a}{b} = \frac{c}{d} $ prove that

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

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

    (2) Now let's consider the case $\displaystyle a\ne 0$ and $\displaystyle c\ne 0$. By a simply manipulation, we found that showing $\displaystyle \frac{ a^2c + ac^2}{b^2d + bd^2} = \frac{(a+c)^3}{(b+d)^3} $ is equivalent to showing $\displaystyle \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.

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

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

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

    Roy
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