If $\frac{a + b}{c} = \frac{b + c}{a} = \frac{c + a}{b} = k$, then $k$ is equal to

### Concept & Addendo Property The Addendo property of proportions states that if a series of ratios are equal, such as $\frac{x}{y} = \frac{p}{q} = \frac{m}{n} = k$, then the sum of all numerators divided by the sum of all denominators is also equal to the same constant $k$, provided the sum of denominators is not zero. ### Step-by-Step Solution 1. **Given:** $\frac{a + b}{c} = \frac{b + c}{a} = \frac{c + a}{b} = k$ 2. Apply the property of equal ratios (Addendo): $$k = \frac{\text{Sum of numerators}}{\text{Sum of denominators}}$$ 3. Add the numerators and denominators together: $$k = \frac{(a + b) + (b + c) + (c + a)}{c + a + b}$$ 4. Combine like terms in the numerator: $$k = \frac{2a + 2b + 2c}{a + b + c}$$ 5. Factor out the $2$: $$k = \frac{2(a + b + c)}{a + b + c}$$ 6. Assuming $a + b + c \neq 0$, the $(a + b + c)$ terms cancel out: $$k = 2$$ ### Exam Strategy & Shortcut Whenever a symmetric cyclic equation like this appears in competitive exams, assume simple non-zero equal values for the variables to solve it instantly. Let $a = 1$, $b = 1$, and $c = 1$. Substitute them into any one of the fractions: $\frac{1 + 1}{1} = \frac{2}{1} = 2$. Therefore, $k = 2$. ### Common Pitfall A common mistake is expanding this into complex simultaneous equations or quadratic forms which wastes immense time. Recognizing the symmetry and either adding them or plugging in $1$ is the optimal path. ### Final Answer Therefore, the correct answer is **$2$**.