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, the equation only makes sense when $a,b,c,d\neq 0$ so you can use this: If
$$\frac{x}{y}=\frac{a}{b}$$
Then
$$\frac{x}{y}=\frac{x+a}{y+b}$$
So your problem solves proving this property, and the only value will be
$$\frac{3(a+b+c+d)}{a+b+c+d}=3$$
But if you get the case $a+b+c+d=0$, then WLOG, $a+b+c=-d$, then you achieve that $a+b+c=-dr$, implying $r=-1$.
Then the only solutions are $r\in \{-1,3\}$, so the solution to the problem would be $3-1=2$.

So to achieve the identity, notice that
$$x(y+b)=y(x+a)$$
$$xy+ab=yx+ya$$
And substracting $xy$ to both sides you get:
$$xb=ya$$
$$\frac{x}{y}=\frac{a}{b}$$
The other side is the same way.
