$\hat{A}=\frac{\stackrel\frown{BC}}{2}=50\div2=25^\circ$ $\hat{BOC}=\stackrel\frown{BC}=50^\circ$ $\hat{AOC}=180-\hat{BOC}=180-50=130^\circ \to \stackrel\frown{AC}=130^\circ$ $\hat{AOE}=\hat{COE}=130\div2=65^\circ \to \stackrel\frown{BD}=65^\circ \to \hat{AOE}=\hat{DOB}=65^\circ$ $\stackrel\frown {AD}=360-(50+130+65)=360-245=115^\circ$