Arithmetic

Ratio, Proportion & Variation

Easy

$\dfrac{(a + b)}{(b + c)} = \dfrac{(c + d)}{(d + a)}$ which of the following statements is always true?

$a + b + c + d = 0$

$a=c,$ or $a+b+c+d = 0$

$a=c$

$a=c,$ and $b=d$

✅ Correct Option: 2

Given:

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

Cross-multiplying:

$(a + b)(d + a) = (c + d)(b + c)$

Expanding both sides:

$ad + a^2 + bd + ab = bc + c^2 + bd + cd$

$bd$ appears on both sides, so cancel it:

$a^2 + ad + ab = c^2 + bc + cd$

Take $a$ common on the LHS and $c$ common on the RHS:

$a^2 + a(b + d) = c^2 + c(b + d)$

Bring everything to one side:

$a^2 - c^2 + a(b + d) - c(b + d) = 0$

Now, $a^2 - c^2 = (a + c)(a - c)$ using the difference of squares identity, and $(b + d)$ is common in the remaining terms:

$(a + c)(a - c) + (b + d)(a - c) = 0$

Take $(a - c)$ common:

$(a - c)\big[(a + c) + (b + d)\big] = 0$

$(a - c)(a + b + c + d) = 0$

When a product of two terms equals zero, at least one of them must be zero. So either:

$a - c = 0 \implies a = c$

$a + b + c + d = 0$

(or both)

Neither condition is guaranteed on its own — but at least one must always hold.

Quick check: let $a = c = 1$ , $b = 2$ , $d = 3$ .

$\dfrac{1 + 2}{2 + 1} = 1$ and $\dfrac{1 + 3}{3 + 1} = 1$ ✅

Here $a = c$ is true, but $a + b + c + d = 7 \neq 0$ . So option (b) is true.

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(a+b)(b+c)=(c+d)(d+a)\dfrac{(a + b)}{(b + c)} = \dfrac{(c + d)}{(d + a)} (b+c)(a+b)​=(d+a)(c+d)​ which of the following statements is always true?

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IPMAT Indore Past Year Questions (Topic-Wise)

$\dfrac{(a + b)}{(b + c)} = \dfrac{(c + d)}{(d + a)}$ which of the following statements is always true?