Júlio Daniel has a square piece of paper with vertices $A, B, C$, and $D$. He first folds this square paper $ABCD$ by bringing the vertices $B$ and $D$ to the diagonal, as shown in the following figure:
![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-07.jpg?height=544&width=1314&top_left_y=550&top_left_x=474)

$\mathrm{E}$ and then, Júlio Daniel brings the vertex $C$ to the vertex $A$, thus obtaining a pentagon, as shown below:
![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-07.jpg?height=538&width=1076&top_left_y=1274&top_left_x=592)

a) Show that the angle $a$ measures $90^{\circ}$.

b) Determine the measure of the angle $b$.
Solution

a) Opening the square of folded paper, one can notice that angle $a$ is the same as the angle at vertex $A$ of the square $A B C D$. Therefore, since all internal angles of a square are equal to $90^{\circ}$, we conclude that $a=90^{\circ}$.

b) Unfolding the paper (only the last fold), we get:
![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-29.jpg?height=362&width=798&top_left_y=2194&top_left_x=731)

The angle $\widehat{A C N}$ is half of the half of the original angle at vertex $C$ of the square. Since the square has all its angles equal to $90^{\circ}$, we conclude that $\widehat{A C N}=$
$\frac{90^{\circ}}{4}=22.5^{\circ}$. By symmetry, the angle $\widehat{C L N}$ is equal to $90^{\circ}$. Since the sum of the internal angles of a triangle is equal to $180^{\circ}$, observing the angles of triangle $C L N$, we have that

$$
90^{\circ}+22.5^{\circ}+\widehat{L N C}=180^{\circ}
$$

Therefore, $\widehat{L N C}=67.5^{\circ}$. Since $\widehat{L N C}+b=180^{\circ}$, we obtain from this that $b=112.5^{\circ}$.