Problem 8.3. On the side $B C$ of rectangle $A B C D$, a point $K$ is marked. Point $H$ on segment $A K$ is such that $\angle A H D=90^{\circ}$. It turns out that $A K=B C$. How many degrees does angle $A D H$ measure if $\angle C K D=71^{\circ}$?

![](https://cdn.mathpix.com/cropped/2024_05_06_b135fe79f2e6ed0cbffag-2.jpg?height=474&width=541&top_left_y=106&top_left_x=462)
Answer: 52.

![](https://cdn.mathpix.com/cropped/2024_05_06_b135fe79f2e6ed0cbffag-2.jpg?height=341&width=507&top_left_y=726&top_left_x=465)

Fig. 5: to the solution of problem 8.3

Solution. Since $B C \| A D$, we get $\angle A D K=\angle C K D=71^{\circ}$ (Fig. 5).

Since $A K=B C=A D$, triangle $A K D$ is isosceles and $\angle A K D=$ $\angle A D K=71^{\circ}$. Then $\angle K A D=180^{\circ}-2 \cdot 71^{\circ}=38^{\circ}$ and $\angle A D H=90^{\circ}-\angle K A D=52^{\circ}$.