Describe how to construct a square inscribed in a triangle such that 2 vertices of the square are on the base of the triangle and the other 2 vertices are one each of the other two sides of the triangle.
Start with a square that is inscribed in the given triangle that has one side on the base of the triangle and one vertex on one of the other sides of the triangle. Then use a dilation from one of the vertices of the triangle so that the fourth vertex of the square comes to lie on the third side of the triangle: this gives you the square that satisfies all requirements.
Here is a picture to illustrate Failure's neat solution. Start by drawing the green square (this is a small square with its base on the base of the triangle and its top left corner somewhere on the left side of the triangle – it doesn't matter where). Then draw the dashed line from the left corner of the triangle to the top right corner of the green square. From the point where this line crosses the right side of the triangle, draw horizontal and vertical lines, as shown in red. You should be able to see (using similar triangles) that the red lines form a square on the base of the triangle, with its top vertices on the other two sides of the triangle, as required.