If \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\), then (a) \(a\) must equal \(c\) (b) \(a+b+c+d\) must equal zero (c) either \(a=c\) or \(a+b+c+d=0\), or both (d) \(a(b+c+d)=c(a+b+d)\)

Starting from the given proportional relationship:

\[
\frac{a+b}{b+c} = \frac{c+d}{d+a}
\]

Multiplying across to eliminate the denominators, we have:

\[
(a + b)(d + a) = (b + c)(c + d)
\]

Expanding both sides:

\[
ad + a^2 + bd + ab = bc + c^2 + cd + bd
\]

Rearranging to group like terms:

\[
a^2 – c^2 + ad – cd + ab – bc = 0
\]

Factoring by grouping, where appropriate, using the difference of squares for \(a^2 – c^2\) and factoring out the common terms in the other parts:

\[
(a – c)(a + c) + (a – c)d + (a – c)b = 0
\]

Factoring \(a – c\) from each term:

\[
(a – c)(a + c + d + b) = 0
\]

For this product to equal zero, at least one of the factors must be zero. Therefore:

\[
a – c = 0 \quad \text{or} \quad a + b + c + d = 0
\]

This means:

– \(a = c\), or
– \(a + b + c + d = 0\), or
– Both conditions could be true in certain scenarios.

Therefore, the correct interpretation is option (c) either \(a = c\) or \(a+b+c+d = 0\), or both.