1964 AHSME Problems/Problem 32: Difference between revisions

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If $\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}$ , then:

$\textbf{(A) }a \text{ must equal }c\qquad\textbf{(B) }a+b+c+d\text{ must equal zero}\qquad$

$\textbf{(C) }\text{either }a=c\text{ or }a+b+c+d=0\text{, or both}\qquad$

$\textbf{(D) }a+b+c+d\ne 0\text{ if }a=c\qquad$

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

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			==Problem==

	If <math>\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}</math>, then:		If <math>\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}</math>, then: