3. In triangle $ABC$, the bisector $BD$ is drawn. It is known that the center of the circumscribed circle of triangle $ABC$ coincides with the center of the circle inscribed in triangle $BCD$. Find the angles of triangle $ABC$.
Answer. $\angle C=36^{\circ}, \angle A=\angle B=72^{\circ}$.

Solution. Let the common center of the circles be $I$, and the points of tangency of the inscribed circle in triangle $BCD$ with sides $CD$, $BC$, and $BD$ be $E$, $F$, and $G$ respectively (see figure). Segments $IE$ and $IF$ are the perpendicular bisectors of sides $AC$ and $BC$ respectively, so $AE=CE$, $BF=CF$. From the properties of tangents to a circle, $BG=BF$, $DG=DE$, $CE=CF$, thus $BD=BG+GD=BF+GD=FC+DE=EC+DE=CD$. Additionally, $BC=2CF=2CE=AC$. Therefore, triangles $BDC$ and $ACB$ are isosceles. Let $\angle BCA=\alpha$, then $\angle CBD=\angle ABD=\alpha$, $\angle CBA=2\alpha$, $\angle CAB=\angle CBA=2\alpha$. Thus, triangle $ABC$ has angles $\alpha$, $2\alpha$, $2\alpha$, from which we find $\alpha$: $5\alpha=180^{\circ}$, $\alpha=36^{\circ}$.

Comment. It is established that triangles $BDC$ and $ACB$ are isosceles - 6 points. For a solution without proof of useful auxiliary statements - 1-3 points.

![](https://cdn.mathpix.com/cropped/2024_05_06_74e32b870097d181f24bg-2.jpg?height=394&width=509&top_left_y=294&top_left_x=1476)