1974 AHSME Problems/Problem 5

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 $\angle BAD$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1974 AHSME Problems/Problem 5

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