The second question:
Take a point B from the center line of the square, so that the distance from it to the two vertices A and C of the square is equal to the side length AC of the square, and then fold it like the above question to get an equilateral triangle. It's easy to find B on the center line. As long as the bottom AC inclines upward from one end A until the other end C falls on the center line, that point is B. Fold AB and CB in half to get the equilateral triangle ABC.
The three angles of an equilateral triangle are equal, which is also called an equilateral triangle or a regular triangle. It has three high lines, which bisect the base vertically and intersect at one point, and divides the equilateral triangle ABC into six right triangles, three isosceles triangles and three quadrangles that can be superimposed. In addition, A'BC' is also an equilateral triangle, and its area is a quarter of ABC's.
I hope I can help you!