Ngle

Let $OE = r$ . Then $OD = OA = 2r$ , so $ED =\sqrt{3}r$ . Also $EB=3r$ . Thus $\Delta EDB$ is a $1-2-\sqrt{3}$ right triangle, so $\angle BDE = 60^{\circ}$ .

Draw segment $OD$ . As a radius of circle with center $O$ , $OD = OA = OB$ .

Since $E$ is the center of $OA$ , and since $OA = OD$ , $OE = \frac{1}{2}OD$ , which means in right $\triangle OED$ $\angle EOD = 60°$ and $\angle ODE = 30°$ .

Since $\angle EOD = 60°$ , $\angle DOB = 180° - 60° = 120°$ . Since $\triangle BOD$ is an isosceles triangle with a vertex angle $\angle DOB = 120°$ , its base angle $\angle ODB = \frac{1}{2}(180° - 120°) = 30°$ .

Since $\angle ODE = 30°$ and $\angle ODB = 30°$ , $\angle BDE = 30° + 30° = \boxed{60°}$ .

AE=EO=r OD=2r DE=3^(1/2)r BE=3r Tan D = BE/DE=3r/((3^(1/2)r)=3^(1/2) D=60°

$AOD \mathrm{ \ is \ an \ equilateral \ triangle} \Rightarrow \angle OAD = 60 \degree \\ \angle ADB = 90 \degree (Thales) \Rightarrow \angle DBE = 30 \degree \Rightarrow \angle BDE = 60 \degree$

Note that $\bigtriangleup AED \cong \bigtriangleup OED$ (SAS criteria), therefore $\bigtriangleup AOD$ is equilateral and, as a consequence: $\angle EAD = 60^{\circ} \Rightarrow \angle BDE = 60^{\circ}$