A terrain ( $ABCD$ ) has a rectangular trapezoidal shape.  The angle in $A$ measures $90^o$. $AB$ measures $30$ m, $AD$ measures $20$ m and $DC$ measures 45 m. This land must be divided into two areas of the same area, drawing a parallel to the $AD$ side . At what distance from $D$ do we have to draw the parallel?
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1. **Identify the given dimensions and shape:**
   - The terrain \(ABCD\) is a rectangular trapezoid.
   - \(\angle A = 90^\circ\).
   - \(AB = 30\) m, \(AD = 20\) m, and \(DC = 45\) m.

2. **Calculate the area of the trapezoid \(ABCD\):**
   - Drop a perpendicular from \(B\) to \(DC\) at point \(K\).
   - Since \(AB\) is perpendicular to \(AD\), \(ABKD\) forms a rectangle.
   - The length of \(BK\) is the same as \(AD\), which is 20 m.
   - The length of \(DK\) is \(DC - AB = 45 - 30 = 15\) m.
   - The area of \(\triangle BKC\) is:
     \[
     [\triangle BKC] = \frac{1}{2} \times DK \times BK = \frac{1}{2} \times 15 \times 20 = 150 \text{ m}^2
     \]
   - The area of rectangle \(ABKD\) is:
     \[
     [ABKD] = AB \times AD = 30 \times 20 = 600 \text{ m}^2
     \]
   - Therefore, the total area of the trapezoid \(ABCD\) is:
     \[
     [ABCD] = [ABKD] + [\triangle BKC] = 600 + 150 = 750 \text{ m}^2
     \]

3. **Determine the area of each of the two equal parts:**
   - Since the land must be divided into two areas of the same area, each part will have an area of:
     \[
     \frac{750}{2} = 375 \text{ m}^2
     \]

4. **Set up the equation for the smaller rectangle:**
   - Let \(x\) be the distance from \(D\) to the parallel line.
   - The area of the smaller rectangle formed by drawing a parallel line to \(AD\) at distance \(x\) from \(D\) is:
     \[
     \text{Area} = 20 \times x
     \]
   - This area must be equal to half the total area of the trapezoid:
     \[
     20x = 375
     \]
   - Solving for \(x\):
     \[
     x = \frac{375}{20} = 18.75 \text{ m}
     \]

The final answer is \(\boxed{18.75 \text{ m}}\)