An interesting result in ratio and proportions

Question

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 : $$ \frac{a}{b}=\frac{c}{d} $$ ??

Answer 1

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 $$ab-cd=0$$

Answer 2

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