1974 AHSME Problems/Problem 5

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Problem

Given a quadrilateral $ABCD$ inscribed in a circle with side $AB$ extended beyond $B$ to point $E$, if $\measuredangle BAD=92^\circ$ and $\measuredangle ADC=68^\circ$, find $\measuredangle EBC$.

$\mathrm{(A)\ } 66^\circ \qquad \mathrm{(B) \ }68^\circ \qquad \mathrm{(C) \  } 70^\circ \qquad \mathrm{(D) \  } 88^\circ \qquad \mathrm{(E) \  }92^\circ$

Solution

Since $ABCD$ is cyclic, opposite angles must sum to $180^\circ$. Therefore, $\angle ADC+\angle ABC=180^\circ$, and $\angle ABC=180^\circ-\angle ADC=180^\circ-68^\circ=112^\circ$. Notice also that $\angle ABC$ and $\angle CBE$ form a linear pair, and so they sum to $180^\circ$. Therefore, $\angle EBC=180^\circ-\angle ABC=180^\circ-112^\circ=68^\circ, \boxed{\text{B}}$. Notice that the answer didn't even depend on $\angle BAD$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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