Proportions manipulation

Question

Knowing that

\begin{equation}\frac{a}{b}=\frac{c}{d}\ \ , \ \ \frac{a'}{b'}=\frac{c'}{d'}\end{equation}

find the condition that the sums

\begin{equation}a+a',\ b+b',\ c+c',\ d+d',\end{equation}

form also a proportion

\begin{equation}\frac{a+a'}{b+b'}=\frac{c+c'}{d+d'}\end{equation}

I tried

\begin{equation}\frac{a}{b}+\frac{a'}{b'}= \frac{c}{d}+\frac{c'}{d'}\end{equation}

\begin{equation}\frac{a}{b}\frac{a'}{b'}= \frac{c}{d}\frac{c'}{d'}\end{equation}

but I did not reach a coclusion

Answer 1

By 'form a proportion' do you mean the following? $$\frac{a+a'}{b+b'}=\frac{c+c'}{d+d'}$$ If so, we can rearrange this to $$(a+a')(d+d')=(b+b')(c+c') \\ ad+a'd+ad'+a'd'=bc+b'c+bc'+b'c'$$ Now note $ad=bc$ and $a'd'=b'c'$, so we require $$a'd+d'a=b'c+bc'$$ Is this what you're looking for? I can't see a good way of expressing this in terms of proportions, if such a way exists.
EDIT: I take that back! There's a cute way: divide by $b'c'$ on the LHS and by $a'd'$ on the RHS (they are the same, so we're ok), and you get $$\frac{a'}{b'}\frac{d}{c'}+\frac{d'}{c'}\frac{a}{b'}=\frac{b'}{a'}\frac{c}{d'}+\frac{c'}{d'}\frac{b}{a'}$$ Which cancels down to $$\frac{d}{d'}+\frac{a}{a'}=\frac{c}{c'}+\frac{b}{b'}$$ Which is sufficiently cute that I'm confident it's what you're looking for.

Answer 2

Let $\frac ab=\frac cd=\lambda$ and $\frac{a'}{b'}=\frac{c'}{d'}=\mu$

Then the relation $\frac{a+a'}{b+b'}=\frac{c+c'}{d+d'}$ becomes $\frac{b\lambda+b'\mu}{b+b'}=\frac{d\lambda+d'\mu}{d+d'}$

This simplifies to $b'd(\mu-\lambda)=bd'(\mu-\lambda)$

So provided $\lambda\neq\mu$, the only condition you need for the stated proportion to hold is $\frac{b}{b'}=\frac{d}{d'}$

However it follows from this that $\frac{a}{a'}=\frac{c}{c'}$

Therefore either of these conditions on it's own is sufficient