Find the value of an angle $x$ $\odot O$ is the circumcircle of an isoceles triangle $\Delta ABC$. 
$AB=AC$, $\measuredangle BAC=20^0$;
$BD$ is a bisects of $\angle ABC$ and intersects $AC$ at $D$.

Find the value of $\measuredangle BDO \qquad$ or prove that $\measuredangle BDO=100^{\circ}$.
 A: It's enough to prove that $OD=DC$.
Indeed, let $BC=a$. 
Hence, by law of sines for $\Delta BDC$ we obtain
$$\frac{DC}{\sin40^{\circ}}=\frac{a}{\sin{120^{\circ}}}$$ or
$$DC=\frac{a\sin40^{\circ}}{\sin{120^{\circ}}}.$$
In another hand, $$a=2OB\sin20^{\circ},$$
which gives $$BO=\frac{a}{2\sin20^{\circ}}.$$
Now, by law of sines for $\Delta BDC$ again we obtain $$BD=\frac{a\sin80^{\circ}}{\sin120^{\circ}}.$$
Thus, by law of cosines for $\Delta OBD$ it's enough to prove that
$$BD^2+BO^2-2BD\cdot BD\cos30^{\circ}=DC^2$$ or
$$\frac{\sin^280^{\circ}}{\frac{3}{4}}+\frac{1}{4\sin^220^{\circ}}-\frac{\sin80^{\circ}}{\sin20^{\circ}}=\frac{\sin^240^{\circ}}{\frac{3}{4}}$$ or
$$16\sin^280^{\circ}\sin^220^{\circ}+3-12\sin80^{\circ}\sin20^{\circ}=16\sin^240^{\circ}\sin^220^{\circ}$$ or
$$4(1-\cos40^{\circ})(\cos80^{\circ}-\cos160^{\circ})+3-6(\cos60^{\circ}-\cos100^{\circ})=0$$ or
$$4\cos80^{\circ}-4\cos160^{\circ}-2\cos120^{\circ}-2\cos40^{\circ}+2\cos120^{\circ}+2\cos200^{\circ}-6\cos80^{\circ}=0$$ or
$$\cos80^{\circ}+\cos40^{\circ}-\cos20^{\circ}=0$$ or
$$2\cos60^{\circ}\cos20^{\circ}=\cos20^{\circ}.$$
Done!
A: Purely geometric solution: 
Since $AB = AC$, the line $AO$ is the orthogonal bisector of segment $BC$. Choose point $E$ on the line $AO$ (and inside the triangle $ABC$) so that $\angle \, BCE = 60^{\circ}$. 

Then, by construction, triangle $BCE$ is equilateral. Thus, $$\angle \, ABE = \angle\, ABC - \angle \, CBE = 80^{\circ} - 60^{\circ} = 20^{\circ}$$
 $$\angle \, ACE = \angle\, ACB - \angle \, BCE = 80^{\circ} - 60^{\circ} = 20^{\circ}$$ However, since $O$ is the circumcenter of $ABC$ we know that $OA - OB = OC$ and $\angle \, BOA = 2 \, \angle \, BCA = 160^{\circ}$ which means that $\angle \, ABO = 10^{\circ}$ and that $$\angle \, EBO = \angle \, ABE - \angle \, ABO = 20^{\circ} - 10^{\circ} = 10^{\circ}$$ Therefore, $BO$ is the angle bisector of $\angle \, ABE$ and by the angle bisector theorem
$$\frac{AO}{OE} = \frac{AB}{BE} = \frac{AB}{BC}$$ because $BE = BC \,$ (triangle $BCE$ is equilateral). However, $BD$ is the angle bisector of angle $\angle \, ABC$ so again by the angle bisector theorem
$$\frac{AD}{DC} = \frac{AB}{BC}$$ Thus
$$\frac{AO}{OE} = \frac{AB}{BC} = \frac{AD}{DC}$$ which by Thales' intercept theorem implies that $DO$ is parallel to $CE$. Consequently, 
$$\angle \, ADO = \angle \, ACE = 20^{\circ}$$ and by angle chasing $\angle \, BDO = 100^{\circ}$.
