3. Given a rectangle ABCD where $\overline{\mathrm{AB}}=2 \overline{\mathrm{BC}}$. On the side $\mathrm{AB}$, a point $\mathrm{K}$ is chosen such that $\angle \mathrm{AKD}=\angle \mathrm{DKC}$. Find this angle.
Solution. Let's draw the height DH of triangle KCD (Fig.1). Triangles KDA and KDH are congruent (right-angled with

![](https://cdn.mathpix.com/cropped/2024_06_05_fb56a3883e438b3946fbg-1.jpg?height=272&width=449&top_left_y=1479&top_left_x=1030)

Fig. 1 common side $\mathrm{KD}$ and $\angle \mathrm{AKD}=\angle \mathrm{DKC}$ ), from which $\overline{\mathrm{DH}}=\overline{\mathrm{DA}}=\frac{\overline{\mathrm{DC}}}{2}$. Since triangle CDH is right-angled, with the leg $\overline{\mathrm{DH}}$ half the hypotenuse $\overline{\mathrm{DC}}$, we get that $\angle \mathrm{HCD}=30^{\circ}$. Then $\angle \mathrm{CKB}=\angle \mathrm{HCD}=30^{\circ}$, and therefore $\angle \mathrm{CKA}=150^{\circ}$. Thus $\angle \mathrm{AKD}=\angle \mathrm{DKC}=75^{\circ}$.