5. A rectangular sheet of paper $ABCD$ was folded as shown in the figure. Find the ratio $DK: AB$, if $C_{1}$ is the midpoint of $AD$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4df8f76a1e5c1d2910b3g-3.jpg?height=728&width=756&top_left_y=201&top_left_x=730)

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# Solution

$A C_{1}=D C_{1}=1 / 2 A D$. Triangles $B C_{1} K$ and $B C K$ are equal, so $B C_{1}=B C$. In the right triangle $A B C_{1}$, the leg $A C_{1}$ is half the hypotenuse $B C_{1}$, which means $\angle A B C_{1}=30^{\circ}$. Therefore, $\angle A C_{1} B=60^{\circ}, \angle D C_{1} K=180^{\circ}-60^{\circ}-90^{\circ}$ $=30^{\circ}$. In the right triangle $K D C_{1}$, the leg opposite the $30^{\circ}$ angle is half the hypotenuse, that is, $D K=1 / 2 C_{1} K=1 / 2 C K$, and since $D K+$ $C K=C D$, then $D K=1 / 3 C D$.

Answer $1: 3$.