In the figure below, \( \angle E = 40^{\circ}\) and arc AB, arc BC, and arc CD all have equal length.
Find the measure of \(\angle ACD\), in degrees.
1. \(\mathbf{ \angle BCE =\ ? }\)
\(\begin{array}{|rcll|} \hline \angle BCE &=& 90^{\circ} - \frac{40^{\circ}}{2} \\ \mathbf{ \angle BCE } &\mathbf{ =}& \mathbf{ 70^{\circ}} \\ \hline \end{array} \)
2. \(\mathbf{ \angle BCA =\ ? }\)
\(\begin{array}{|rcll|} \hline \angle BCE + 2\cdot \angle BCA &=& 180^{\circ} \\ \angle BCA &=& \frac{180^{\circ}- \angle BCE} {2} \\ \angle BCA &=& 90^{\circ}- \frac{\angle BCE} {2} \quad & | \quad \angle BCE = 70^{\circ} \\ \angle BCA &=& 90^{\circ}- \frac{70^{\circ}} {2} \\ \angle BCA &=& 90^{\circ}- 35^{\circ} \\ \mathbf{ \angle BCA} &\mathbf{ =}& \mathbf{ 55^{\circ}} \\ \hline \end{array} \)
3. \(\mathbf{ \angle ACD =\ ? }\)
\(\begin{array}{|rcll|} \hline \angle ACD &=& \angle BCE - \angle BCA \\ \angle ACD &=& 70^{\circ} - 55^{\circ} \\ \mathbf{ \angle ACD} &\mathbf{ =}& \mathbf{ 15^{\circ} } \\ \hline \end{array}\)
+9491 
