## Task A-2.5.

Given is a square $A B C D$. Let $E$ be a point on the ray $A B$ such that $\varangle A E D=27^{\circ}$. The segments $\overline{A C}$ and $\overline{D E}$ intersect at point $S$. Determine the measure of the angle $\varangle B S E$.
## First Solution.

From the right triangle $A E D$, we have $\varangle E D A=90^{\circ}-\varangle A E D=90^{\circ}-27^{\circ}=63^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_21f39afeb1dfc245d841g-10.jpg?height=457&width=788&top_left_y=2036&top_left_x=565)

The angles $\varangle B A S$ and $\varangle S A D$ lie along the diagonal of the square, so they measure $45^{\circ}$. Also, $|A B|=|A D|$. Finally, the side $\overline{A S}$ is common to triangles $A B S$ and $A D S$. From all this, we conclude that these triangles are congruent by the S-A-S (Side-Angle-Side) congruence theorem.

Thus, we have $\varangle A B S=\varangle S D A=\varangle E D A=63^{\circ}$.

Finally, in triangle $S B E$, we have

$$
\begin{aligned}
\varangle B S E & =180^{\circ}-\varangle S B E-\varangle B E S \\
& =180^{\circ}-\left(180^{\circ}-\varangle A B S\right)-\varangle A E D \\
& =\varangle A B S-\varangle A E D=63^{\circ}-27^{\circ}=36^{\circ}
\end{aligned}
$$