B2. In the figure, you see an equilateral triangle $A B C$ and points $D$ and $E$ on sides $B C$ and $A C$. If you fold the triangle along the line $D E$, then vertex $C$ lands exactly on line $A B$, at point $C^{\prime}$. It is given that $\angle D C^{\prime} B=90^{\circ}$.

How large is $\angle D E C^{\prime}$?

Note: the picture is not drawn to scale.

![](https://cdn.mathpix.com/cropped/2024_04_17_8b0e5b2f470f23b328a2g-1.jpg?height=377&width=483&top_left_y=1345&top_left_x=1346)
B2. $45^{\circ}$

First, we note that triangles $C D E$ and $C^{\prime} D E$ are mirror images of each other and thus have equal angles. In particular, $\angle D C^{\prime} E = \angle D C E = 60^{\circ}$. It also follows that $\angle C E D = \angle D E C^{\prime}$, see the figure.

From $\angle A C^{\prime} B = 180^{\circ}$, we now get that $\angle A C^{\prime} E = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$. The sum of the angles in triangle $A C^{\prime} E$ is $180^{\circ}$, and thus we find $\angle A E C^{\prime} = 180^{\circ} - 60^{\circ} - 30^{\circ} = 90^{\circ}$. From $\angle A E C = 180^{\circ}$, we now get that $\angle C E C^{\prime} = 180^{\circ} - 90^{\circ} = 90^{\circ}$.

We conclude that $\angle D E C^{\prime} = \frac{1}{2} \angle C E C^{\prime} = 45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ff229bb29d2323b894dcg-1.jpg?height=432&width=477&top_left_y=1029&top_left_x=1372)