<s>[INST] An interesting result in ratio and proportions If 
$$
\frac{a}{b}=\frac{c}{d}=k
$$
then 
$$
\frac{a+c}{b+d}=k
$$
Also
$$
\frac{a^2}{b^2}=\frac{c^2}{d^2}=k^2
$$
And
$$
\left( \frac{a+c}{b+d} \right)^2=k^2
$$
Also 
$$
\frac{a^2 + c ^2}{b^2+d^2}=k^2
$$
Hence
$$
\frac{a^2 + c ^2}{b^2+d^2}= \left( \frac{a+c}{b+d} \right)^2
$$
My question is... if the reverse is true. That is...
if 
$$
\frac{a^2 + c ^2}{b^2+d^2}= \left( \frac{a+c}{b+d} \right)^2
$$
Then can we assume that :
$$
\frac{a}{b}=\frac{c}{d}
$$
??
 [/INST] It is clear that the truth of  $$ \frac{a^2+c^2}{b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}$$
does not change if we swap $b\leftrightarrow d$, but the truth of $\frac ab=\frac cd$ may well change
</s>