## Task 3 - 080623

Above the side $C D$ of a square $A B C D$ with $A B=4 \mathrm{~cm}$, an equilateral triangle $\triangle D C E$ is to be constructed such that the square and the triangle share the side $C D$.

The point $E$ of the triangle $\triangle D C E$ is to be located outside the square $A B C D$. Connect $E$ with $A$ and with $B$!

Calculate the size of the angle $\angle A E B$!
In the triangle $\triangle A E D$ it holds: (see figure) $A D=D E$ (by construction)

(1) This implies $\angle D A E=\angle A E D$ (as base angles). Furthermore:

(2) $\angle E D A=\angle E D C+\angle C D A=150^{\circ}$; $\angle A B C$ denotes the measure of the angle $\left.\angle A B C\right)$ since $\angle E D C=$ $60^{\circ}$ (angle in an equilateral triangle) and $\angle C D E=90^{\circ}$ (angle in a square).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0183.jpg?height=408&width=286&top_left_y=247&top_left_x=888)

From (1), (2), and $\angle A E D+\angle E D A+\angle D A E=180^{\circ}$ (by the angle sum property), it follows that $\angle A E D=15^{\circ}$. Similarly, $\angle B E C=15^{\circ}$, thus $\angle A E B=\angle D E C-\angle A E D-\angle B E C=30^{\circ}$.