<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] if we factorizing $$\frac{a^2+c^2}{b^2+d^2}-\left(\frac{a+c}{b+d}\right)^2=\frac{2 (a d-b c) (a b-c d)}{(b+d)^2 \left(b^2+d^2\right)}=0$$ and we get $$ad-bc=0$$ or $$ab-cd=0$$
</s>