1. ## 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

$\displaystyle \frac{ a^2c + ac^2}{b^2d + bd^2} = \frac{(a+c)^3}{(b+d)^3}$

2. Originally Posted by Percent09
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