In the adjoining figure $\measuredangle E=40^\circ$ and arc $AB$, arc $BC$, and arc $CD$ all have equal length. Find the measure of $\measuredangle ACD$.
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }\left(\frac{45}{2}\right)^\circ\qquad \textbf{(E) }30^\circ$
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, $\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ$. $\theta = 55^\circ$. $\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{\textbf{(B) }15^\circ}$.