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Deduction from a fraction

Source: AHSME 1964 Problem 32

January 13, 2014
AMC

Problem Statement

If a+bb+c=c+dd+a\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}, then:
<spanclass=′latex−bold′>(A)</span> a must equal c<spanclass=′latex−bold′>(B)</span> a+b+c+d must equal zero <span class='latex-bold'>(A)</span>\ a\text{ must equal }c \qquad <span class='latex-bold'>(B)</span>\ a+b+c+d\text{ must equal zero }\qquad
<spanclass=′latex−bold′>(C)</span> either a=c or a+b+c+d=0, or both <span class='latex-bold'>(C)</span>\ \text{either }a=c\text{ or }a+b+c+d=0,\text{ or both} \qquad
<spanclass=′latex−bold′>(D)</span> a+b+c+d≠0 if a=c<spanclass=′latex−bold′>(E)</span> a(b+c+d)=c(a+b+d) <span class='latex-bold'>(D)</span>\ a+b+c+d\neq 0\text{ if }a=c \qquad <span class='latex-bold'>(E)</span>\ a(b+c+d)=c(a+b+d)
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