4.25. The centers of the circumcircle and the incircle of a triangle are symmetric with respect to one of the sides of the triangle. Find the angles of the triangle.
4.25. Let $\angle B D C=x$; then $\angle B O C=$ $=2 x$ (Fig. 4.22). Therefore, $\angle O B C=$ $=90^{\circ}-x$, i.e., $\angle O_{1} B C=90^{\circ}-x$ (due to the symmetry of points $O$ and $O_{1}$ relative to side $B C$). Let's write the sum of the angles of $\triangle A B C$: $180^{\circ}-x+180^{\circ}-2 x+180^{\circ}-2 x=540^{\circ}-$ $-5 x$; then $540^{\circ}-5 x=180^{\circ}$, from which

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-584.jpg?height=419&width=406&top_left_y=1427&top_left_x=747)

Fig. 4.22
$x=72^{\circ}$. Finally, we get $\angle B A C=180^{\circ}-72^{\circ}=108^{\circ}, \angle C B A=$ $=\angle B C A=\frac{1}{2}\left(180^{\circ}-108^{\circ}\right)=36^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-585.jpg?height=320&width=342&top_left_y=290&top_left_x=46)

Fig. 4.23
Answer: $108^{\circ}, 36^{\circ}$ and $36^{\circ}$.