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Difference between revisions of "1977 AHSME Problems/Problem 9"

(Created page with "==Solution== Solution by e_power_pi_times_i If arcs <math>AB</math>, <math>BC</math>, and <math>CD</math> are congruent, then <math>\measuredangle ACB = \measuredangle BDC =...")
(No difference)

Revision as of 12:28, 18 November 2016

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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