2. In triangle $ABC$, the bisector $BD$ is drawn. It is known that the center of the circumcircle around $\triangle ABC$ coincides with the center of the circle inscribed in $\triangle BCD$. Find the angles of $\triangle ABC$.
Solution. Let the common center of the circles be denoted by $I$, and the points of tangency of the incircle of $\triangle BCD$ with the sides $CD, BC$, and $BD$ be denoted by $E, F, G$ respectively (Fig. 1). The segments $IE$ and $IF$ are the perpendicular bisectors of the sides $AC$ and $BC$ respectively, so $AE = CE$ and $BF = CF$. From the properties of tangents to a circle, we have $BG = BF$, $CF = CE$, and $DG = DE$, therefore

$$
BD = BG + DG = BF + DG = CF + DE = CE + DE = CD
$$

Also, we have that

$$
BC = 2CF = 2CE = AC
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ef3d520eb62e6311272ag-1.jpg?height=586&width=851&top_left_y=1917&top_left_x=797)

Fig. 1

$$
\begin{gathered}
\angle BAC = \angle ABC = 2 \angle DBC = 2 \angle BCA \Rightarrow \\
180^{\circ} = \angle BAC + \angle ABC + \angle BCA = 5 \angle BCA \Rightarrow \\
\angle BCA = 36^{\circ}, \angle BAC = \angle ABC = 2 \angle BCA = 72^{\circ}
\end{gathered}
$$

Answer: $\angle A = \angle B = 72^{\circ}, \angle C = 36^{\circ}$.