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 $\boxed{18.75 \text{ m}}$