A trapezoid $ABCD$ with bases $BC = a$ and $AD = 2a$ is drawn on the plane. Using only with a ruler, construct a triangle whose area is equal to the area of the trapezoid. With the help of a ruler you can draw straight lines through two known points.

(Rozhkova Maria)
1. **Extend the bases of the trapezoid**: Extend the lines containing the bases \(BC\) and \(AD\) of the trapezoid \(ABCD\) so that they intersect at a point. Let's call this point \(P\).

2. **Construct a new base**: On one of the extended lines, construct a segment \(PQ\) such that \(PQ = 3a\). This can be done by marking off three segments each of length \(a\) along the line.

3. **Connect endpoints to form a triangle**: Connect the endpoints \(P\) and \(Q\) to any point \(R\) on the other extended line. This forms a triangle \(PQR\).

4. **Calculate the area of the triangle**: The area of the triangle \(PQR\) can be calculated using the formula for the area of a triangle:
   \[
   \text{Area of } \triangle PQR = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Here, the base \(PQ = 3a\) and the height \(h\) is the perpendicular distance from \(R\) to the line containing \(PQ\).

5. **Compare with the area of the trapezoid**: The area of the trapezoid \(ABCD\) is given by:
   \[
   \text{Area of trapezoid } ABCD = \frac{1}{2} \times (BC + AD) \times h = \frac{1}{2} \times (a + 2a) \times h = \frac{3}{2} \times a \times h
   \]
   The area of the triangle \(PQR\) is:
   \[
   \text{Area of } \triangle PQR = \frac{1}{2} \times 3a \times h = \frac{3}{2} \times a \times h
   \]
   Therefore, the area of the triangle \(PQR\) is equal to the area of the trapezoid \(ABCD\).

Conclusion:
By constructing a triangle \(PQR\) with base \(PQ = 3a\) and height \(h\) (the same height as the trapezoid), we have created a triangle whose area is equal to the area of the trapezoid.

The final answer is \( \boxed{ \triangle PQR } \) with base \(PQ = 3a\) and height \(h\).