If $\dfrac{a}{b+c} = \dfrac{b}{c+a} = \dfrac{c}{a+b}$, then each ratio is equal to:

1. $\dfrac{1}{2}$

2. −1

3. either $\dfrac{1}{2}$ or −1

4. neither $\dfrac{1}{2}$ nor −1

Given,

$\dfrac{a}{b+c} = \dfrac{b}{c+a} = \dfrac{c}{a+b} = k$

From the first relation: a = k(b + c) ….(1)

From the second relation: b = k(c + a) ….(2)

From the third relation: c = k(a + b) ….(3)

Adding (1), (2) and (3):

a + b + c = k(b + c) + k(c + a) + k(a + b)

a + b + c = k((b + c) + (c + a) + (a + b))

a + b + c = k(b + c + c + a + a + b)

a + b + c = k(a + a + b + b + c + c)

a + b + c = k(2a + 2b + 2c)

a + b + c = 2k(a + b + c)

If a + b + c ≠ 0, then,

$2k =\dfrac{(a + b + c)}{(a + b + c)}$

2k = 1

$k = \dfrac{1}{2}.$

If a + b + c = 0, then the equations will be,

a + b = -c ….(4)

b + c = -a ….(5)

c + a = -b ….(6)

Subtituting values in $\dfrac{a}{b+c} = \dfrac{b}{c+a} = \dfrac{c}{a+b}$

$\dfrac{a}{-a} = \dfrac{b}{-b} = \dfrac{c}{-c} = k$

k = -1

Hence, Option 3 is the correct option.