1977 AHSME Problems/Problem 9

Solution

Solution by e_power_pi_times_i

If arcs $AB$ , $BC$ , and $CD$ are congruent, then $\measuredangle ACB = \measuredangle BDC = \measuredangle CBD = \theta$ . Because $ABCD$ is cyclic, $\measuredangle CAD = \measuredangle CBD = \theta$ , and $\measuredangle ADB = \measuredangle ACB = \theta$ . Then, $\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ$ . $\theta = 55^\circ$ . $\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{\textbf{(B) }15^\circ}$ .

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1977 AHSME Problems/Problem 9

Solution