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$\odot O$ 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 $D$. enter image description here

Find the value of $\measuredangle BDO \qquad$ or prove that $\measuredangle BDO=100^{\circ}$.

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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 $\angle \, BCE = 60^{\circ}$.

enter image description here

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 $\angle \, BDO = 100^{\circ}$.

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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!

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  • $\begingroup$ Cool! Someone else uses the result of this problem to prove the same thing. While another problem is also difficult $\endgroup$ – LCFactorization Aug 14 '17 at 11:24
  • $\begingroup$ @LCFactorization I tried the second problem. It's indeed interesting enough. $\endgroup$ – Michael Rozenberg Aug 14 '17 at 11:34
  • $\begingroup$ For the problem in this post, there is another theorem to use: $$\dfrac{\sin\measuredangle ABO}{\sin\measuredangle OBD}\cdot \dfrac{\sin x}{\sin(120^0-x)}=1$$ $\endgroup$ – LCFactorization Aug 14 '17 at 11:45
  • $\begingroup$ @LCFactorization You are right! It's easier. $\endgroup$ – Michael Rozenberg Aug 14 '17 at 11:53

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