**Problem**

The base AB of two equilateral triangles ABC and ABD with side 2a lies along the x-axis such that the midpoint of AB is at the origin. Find the coordinates of two vertices C and D of the triangle.

**Solution**

Let us first draw the diagram given in the problem statement.


As given in the problem statement, AB is the base of two equilateral triangles ABC and ABD with side 2a. We can see from the diagram that the points C and D lie on the circles with centers at A and B respectively and radius 2a.

Let us now find the coordinates of point C. We know that the point A lies on the x-axis and the midpoint of AB is at the origin. Therefore, the x-coordinate of point C is a. The y-coordinate of point C can be found by considering the right-angled triangle OAC. In this triangle, the hypotenuse OA has length 2a (since A is the center of the circle with radius 2a), and the side OA has length a (since A lies on the x-axis). Therefore, the y-coordinate of point C is √3a.

So, the coordinates of point C are (a, √3a).

Similarly, we can find the coordinates of point D. Since the point B lies on the x-axis and the midpoint of AB is at the origin, the x-coordinate of point D is -a. To find the y-coordinate of point D, we consider the right-angled triangle OBD. In this triangle, the hypotenuse OB has length 2a (since B is the center of the circle with radius 2a), and the side BD has length a (since B lies on the x-axis). Therefore, the y-coordinate of point D is -√3a.

So, the coordinates of point D are (-a, -√3a).

Hence, the coordinates of the vertices C and D of the equilateral triangles ABC and ABD are (a, √3a) and (-a, -√3a) respectively.

**Answer**

The coordinates of the vertices C and D of the equilateral triangles ABC and ABD are (a, √3a) and (-a, -√3a) respectively.