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Triangle $\triangle ABC$ is inscribed in a circle. $D$ is a point on $AC$. $BD$ is angle bisector of $\angle B$. $O$ is the center of the circle then find $\angle ADO$ if $\angle A=20°$ and $AB=AC.$

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closed as off-topic by José Carlos Santos, Thomas Shelby, Yanior Weg, Leucippus, GNUSupporter 8964民主女神 地下教會 Apr 22 at 23:04

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    $\begingroup$ Interesting enough. What is the answer? $\endgroup$ – MarianD Apr 22 at 2:36
  • $\begingroup$ The calculation is tedious, but possible the $\angle BOC=40^{\circ}$ and lots of cosine- and sine-theorems. A GeoGebra construction results in $20^{\circ}$ again. This only happens for $20^{\circ}$ and is therefore quite remarkable. $\endgroup$ – Strichcoder Apr 22 at 3:07
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Let $P$ be the point on the circumcircle of $ABC$ such that $BOP$ is equilateral. So $\angle POC = 20^\circ$ and then $\angle CBP = 10^\circ$. Note that $\angle CPB=20^\circ$ Also note that $\triangle OXC \cong \triangle PCB$, where $X$ is the intersection of $OP$ and $AC$, therefore $XC=BC$. By the Bisector Theorem we have $$\frac{AB}{BC} = \frac{AD}{DC} = \frac{DC}{XD}.$$ But $\triangle ABC \sim \triangle OCP$, so $$\frac{OC}{CP} = \frac{AB}{BC},$$ so $$\frac{DC}{XD} = \frac{OC}{CP} = \frac{OC}{OX}.$$ Therefore, by Bisector Theorem we've finished.

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