$\underline{\text { Folklore }}$

For some numbers $a, b, c$ and $d$, different from zero, the equality holds: $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$. Determine the sign of the number $a c$.
The equality is transformed into the form $(a d+b c)(b+d)=(a+c) b d$. By expanding the brackets and canceling equal terms, we get $a d^{2}+b^{2} c=0$. From this, it is clear that the numbers $a d^{2}$ and $b^{2} c$ (and therefore the numbers $a$ and $c$) are of opposite signs. Therefore, $ac < 0$.

Answer

$ac < 0$.