## Task $7 / 82$

Let $a ; b ; c ; d$ be four non-zero real numbers for which the equations

$$
\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=d
$$

hold. Determine all possible values of $d$.
From

$$
\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=d
$$

follow the equations

$$
a+b=c d \quad ; \quad a+c=b d \quad ; \quad b+c=a d
$$

By adding these equations side by side, we obtain $2(a+b+c)=d(a+b+c)$.

There are now two possibilities:

1. $a+b+c=0$. It follows that $a+b=-c ; a+c=-b ; b+c=-a$ and thus

$$
\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=\frac{-c}{c}=\frac{-b}{b} \frac{-a}{a}=1=d
$$

2. $a+b+c \neq 0$. Then it immediately follows that $d=2$. The possible values for $d$ are therefore $d_{1}=-1$ and $d_{2}=2$.