# Difference between revisions of "1964 AHSME Problems/Problem 32"

## Problem

If $\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}$, then: $\textbf{(A) }a \text{ must equal }c\qquad\textbf{(B) }a+b+c+d\text{ must equal zero}\qquad$ $\textbf{(C) }\text{either }a=c\text{ or }a+b+c+d=0\text{, or both}\qquad$ $\textbf{(D) }a+b+c+d\ne 0\text{ if }a=c\qquad$ $\textbf{(E) }a(b+c+d)=c(a+b+d)$

## Solution

Cross-multiplying gives: $(a+b)(a+d) = (b+c)(c+d)$ $a^2 + ad + ab + bd = bc + bd + c^2 + cd$ $a^2 + ad + ab - bc - c^2 - cd = 0$ $a(a + b + d) - c(b+c+d)= 0$

This looks close to turning into option C, but we don't have a $c$ term in the first parentheses, and we don't have an $a$ term in the second parentheses to allow us to complete the factorization. However, if we both add $ac$ and subtract $ac$ on the LHS, we get: $a(a + b + d) + ac - c(b+c+d) - ca= 0$ $a(a+b+d +c) - c(b+c+d+a) = 0$ $(a-c)(a+b+c+d) = 0$

This is equivalent to $\boxed{\textbf{(C)}}$

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