9.8. In an acute-angled triangle $A B C$, the bisector $B L$ is drawn. The circumcircle of triangle $A B L$ intersects side $B C$ at point $D$. It turns out that point $S$, symmetric to point $C$ with respect to the line $D L$, lies on side $A B$ and does not coincide with its endpoints. What values can $\angle A B C$ take?

(B. Obukhov, zhori)
Answer: $60^{\circ}$.

First solution. From the symmetry, triangles $C L D$ and $S L D$ are equal, so $D S = D C$, $\angle C D L = \angle S D L$, and $\angle D L C = \angle D L S$. Since the quadrilateral $A L D B$ is inscribed in a circle, we have $\angle B A L = \angle L D C$ (see Fig. 1). The chords $A L$ and $D L$ of this circle subtend equal angles, so $A L = D L$.

We mark off a segment $A S^{\prime} = D S = D C$ on the ray $A B$. Then triangles $S^{\prime} L A$ and $S L D$ are equal by two sides ($A S^{\prime} = D S$, $A L = D L$) and the angle between them. Therefore, $L S^{\prime} = L S$ and $\angle A S^{\prime} L = \angle D S L$. If points $S^{\prime}$ and $S$ do not coincide, then in the isosceles triangle $L S S^{\prime}$, the angles at the base $S S^{\prime}$ are equal, so $\angle A S^{\prime} L = \angle B S L$. Thus, $\angle D S L = \angle B S L$, which means $D$ lies on $A B$; this is impossible.

Therefore, $S^{\prime} = S$, and $\angle A L S = \angle D L S = \angle D L C$. The sum of these three angles is $180^{\circ}$, so they are each $60^{\circ}$. Hence, $\angle A B C = 180^{\circ} - \angle A L D = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e8607f50b48dc7c8d053g-08.jpg?height=444&width=489&top_left_y=442&top_left_x=224)

![](https://cdn.mathpix.com/cropped/2024_05_06_e8607f50b48dc7c8d053g-08.jpg?height=402&width=521&top_left_y=437&top_left_x=743)

Fig. 2

Second solution. Let $\Omega$ be the circumcircle of triangle $B C S$ (see Fig. 2). In this triangle, the line $B L$ is the angle bisector of angle $B$, and the line $L D$ is the perpendicular bisector of side $C S$. These lines are distinct; both pass through the midpoint of the arc $C S$ of circle $\Omega$ that does not contain $B$. Therefore, their intersection point $L$ is this midpoint, i.e., $L \in \Omega$.

Since the quadrilateral $A L D B$ is inscribed, we have $\angle A B C = \angle D L C = \beta$. From symmetry, $\angle D L C = \angle D L S = \beta$. Now from circle $\Omega$, we get $180^{\circ} = \angle C B S + \angle S L C = \beta + \angle D L S + \angle D L C = 3 \beta$, from which $\beta = 60^{\circ}$.

Remark. There are other solutions. For example, one can notice that point $L$ is the excenter of triangle $B D S$ (since $B L$ is the angle bisector of the internal angle of this triangle, and $D L$ is the angle bisector of its external angle). This leads to the calculation $\angle A B C = \angle D L C = \angle D L S = 90^{\circ} - \angle A B C / 2$, from which the answer follows.

One can also consider the point $T$, symmetric to point $C$ with respect to $L B$, as well as the projections $S^{\prime}$ and $T^{\prime}$ of point $C$ onto $L D$ and $L B$ respectively. From the inscribed quadrilaterals $A B D L$ and $L T^{\prime} S^{\prime} C$, we have $\angle A B C = \angle D L C = \angle S^{\prime} T^{\prime} C = \angle S T C = \angle B T C$. But triangle $B T C$ is isosceles, so $\angle B T C = 90^{\circ} - \angle A B C / 2$.

Comment. Without additional considerations, it is claimed that triangles $A L S$ and $D L S$ are equal (by two sides and the angle $\angle L A S = \angle L D S$ not between them) - no more than 3 points for the problem.