How to get the value of the expression? I have this statement:

If $\frac{a}{b+c+d} + \frac{b}{a+c+d} + \frac{c}{a+b+d} + \frac{d}{a+b+c} = 1$
Find the value of $\frac{a^2}{b+c+d} + \frac{b^2}{a+c+d} + \frac{c^2}{a+b+d} + \frac{d^2}{a+b+c}$

My attempt was:
I will denote $k = a+b+c+d$ and $\triangle_n = k-n$
Then,
$\frac{a}{\triangle_a}+\frac{b}{\triangle_b}+\frac{c}{\triangle_c}+\frac{d}{\triangle_d}=1$
Now, multiply by $\frac{a}{a},\frac{b}{b}$ and so on, thus:
$\frac{a^2}{a\triangle_a}+\frac{b^2}{b\triangle_b}+\frac{c^2}{c\triangle_c}+\frac{d^2}{d\triangle_d}=1$
and I do not know how to get more information.
Any hint for this specific problem is aprecciated, and any hint for this type of problem is appreciated too.

$$a+b+c+d=(a+b+c+d)\sum_{cyc}\frac{a}{b+c+d}=$$
$$=\sum_{cyc}\frac{a^2+a(b+c+d)}{b+c+d}=a+b+c+d+\sum_{cyc}\frac{a^2}{b+c+d},$$
which says $$\sum_{cyc}\frac{a^2}{b+c+d}=0.$$
In the full writing the last equality it's:
$$\frac{a^2}{b+c+d}+\frac{b^2}{c+d+a}+\frac{c^2}{d+a+b}+\frac{d^2}{a+b+c}=0.$$
