Rectangular sheet of paper $ABCD$ is folded as shown in the figure. Find the rato  $DK: AB$, given that $C_1$ is the midpoint of $AD$.
[img]https://3.bp.blogspot.com/-9EkSdxpGnPU/W6dWD82CxwI/AAAAAAAAJHw/iTkEOejlm9U6Dbu427vUJwKMfEOOVn0WwCK4BGAYYCw/s400/Yasinsky%2B2017%2BVIII-IX%2Bp1.png[/img]
1. **Identify the given information and the goal:**
   - We have a rectangular sheet of paper \(ABCD\).
   - \(C_1\) is the midpoint of \(AD\).
   - We need to find the ratio \(DK:AB\).

2. **Analyze the given figure and the fold:**
   - When the paper is folded, point \(C\) is brought to point \(C_1\).
   - Since \(C_1\) is the midpoint of \(AD\), \(C_1\) divides \(AD\) into two equal parts.

3. **Determine the properties of \(\Delta BC_1C\):**
   - Given that \(\Delta BC_1C\) is equilateral, all sides are equal, and all angles are \(60^\circ\).

4. **Calculate the angles involved:**
   - Let \(\angle KC_1D = x\).
   - Since \(\Delta BC_1C\) is equilateral, \(\angle BC_1C = 60^\circ\).
   - \(\angle C_1CD = 45^\circ - \frac{x}{2}\) because \(C_1\) is the midpoint and the fold creates a \(45^\circ\) angle with the horizontal.

5. **Solve for \(x\):**
   - Given \(\angle C_1CD = 30^\circ\), we have:
     \[
     45^\circ - \frac{x}{2} = 30^\circ
     \]
     \[
     45^\circ - 30^\circ = \frac{x}{2}
     \]
     \[
     15^\circ = \frac{x}{2}
     \]
     \[
     x = 30^\circ
     \]

6. **Determine the ratio \(\frac{DK}{KC_1}\):**
   - Since \(\Delta BC_1C\) is equilateral, \(BC_1 = C_1C\).
   - Given that \(\angle KC_1D = 30^\circ\), \(\Delta DKC_1\) is a 30-60-90 triangle.
   - In a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\).

7. **Calculate the ratio \(DK:AB\):**
   - Since \(DK\) is opposite the \(30^\circ\) angle, and \(KC_1\) is opposite the \(60^\circ\) angle, we have:
     \[
     \frac{DK}{KC_1} = \frac{1}{\sqrt{3}}
     \]
   - Given that \(KC_1 = \frac{1}{2} AB\) (since \(C_1\) is the midpoint of \(AD\) and \(AD = AB\)), we have:
     \[
     \frac{DK}{AB} = \frac{1}{2\sqrt{3}} = \frac{1}{3}
     \]

The final answer is \(\boxed{1:3}\).