## Task 3 - 200733

Let $S$ be the vertex of an acute angle, whose sides are denoted by $s_{1}$ and $s_{2}$. It is assumed that on the ray $s_{1}$ there are two points $A, B$ different from each other and from $S$, and that on the ray $s_{2}$ there are three points $C, D, E$ different from each other and from $S$, such that the following holds:

The points $S, A, B$ are arranged on $s_{1}$ in this order; the points $S, C, D, E$ are arranged on $s_{2}$ in this order; it is $S C=C A=A D=D B=B E$ and $S B=S E$.

Determine the size $\alpha$ of the angle $\angle B S E$ from these assumptions!
Given the conditions:

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0449.jpg?height=251&width=688&top_left_y=1222&top_left_x=250)

\(\angle C A S = \angle A S C = \alpha\) because \(\triangle A C S\) is isosceles with \(A C = C S\),

\(\angle A C D = \angle C A S + \angle A S C = 2 \alpha\) as the exterior angle of \(\triangle A C S\),

\(\angle A D C = \angle A C D = 2 \alpha\) because \(\triangle A C D\) is isosceles with \(A C = A D\),

\(\angle B A D = \angle A D C + \angle A S C = 3 \alpha\) as the exterior angle of \(\triangle A D S\),

\(\angle A B D = \angle D A B = 3 \alpha\) because \(\triangle A B D\) is isosceles with \(A D = B D\),

\(\angle B D E = \angle A B D + \angle A S C = 4 \alpha\) as the exterior angle of \(\triangle B D S\),

\(\angle B E D = \angle B D E = 4 \alpha\) because \(\triangle B D E\) is isosceles with \(B D = B E\),

\(\angle E B S = \angle B E D = 4 \alpha\) because \(\triangle B E S\) is isosceles with \(B S = E S\),

and from the angle sum in \(\triangle B E S\):

\[
\alpha + 4 \alpha + 4 \alpha = \angle B S E + \angle B E D + \angle E B S = 180^{\circ} \quad \text{and thus} \quad \alpha = 20^{\circ}
\]