# Geometry triangle inside circle [closed]

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 $$AB=AC.$$

## 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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• Interesting enough. What is the answer? – MarianD Apr 22 at 2:36
• 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. – Strichcoder Apr 22 at 3:07

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.