We have $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$. 
If $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$ , find the sum of all the possible values of $r$ .

What I Tried :- First of all, when $a = b = c = d$ , the $4$ equations hold and we get $r = 3$ as $1$ solution .
For other solutions I simplified to get $6$ expressions as :-
$b^2 + bc + bd = a^2 + ac + ad$
$bc + c^2 + cd = a^2 + ab + ad$
$bd + cd + d^2 = a^2 + ab + ac$
$ac + c^2 + cd = ab + b^2 + bd$
$ad + cd + d^2 = ab + b^2 + bc$
$ad + bd + d^2 = ac + bc + c^2$
Now I have no idea how to start finding solutions for $r$ from here . Can anyone help?

First of all $$\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$$ gives you $$b + c + d=ar\\a + c + d=br\\a + b + d=cr\\a + b + c=dr$$ summing all these we get $$3(a+b+c+d)=r(a+b+c+d)$$ Now if $(a+b+c+d)\neq0$ then we must have $r=3$.
Now if $(a+b+c+d)=0$, then $r=\frac{a+b+c}{d}\implies r=\frac{a+b+c+d-d}{d}=\frac{-d}{d}=-1$ (For clarity: since we are writing $a,b,c,d$ in the denominators, we must need $a,b,c,d$ to be non-zero.). Therefore the sum of all possible values of $r$ is $3-1=2$.
