Prove that the figure ABCD whose vertices are the points (-2,0),0,5),(2,0) and (0,-5) is not a square?

To prove that the figure ABCD is not a square, we need to show that the lengths of its sides are not all equal and that its angles are not all right angles.

1. Calculate the lengths of the sides:
- Side AB: √((-2-0)^2 + (0-5)^2) = √(4 + 25) = √29
- Side BC: √((0-5)^2 + (5-0)^2) = √(25 + 25) = √50
- Side CD: √((5-2)^2 + (0+5)^2) = √(9 + 25) = √34
- Side DA: √((2+2)^2 + (0+0)^2) = √(4 + 0) = 2

Since the lengths of the sides are not all equal, the figure ABCD is not a square.

2. Calculate the angles:
- Angle ABC: arctan((5-0)/(0-5)) = arctan(-1) = -45 degrees
- Angle BCD: arctan((0-5)/(5-0)) = arctan(-1) = -45 degrees
- Angle CDA: arctan((0+5)/(2+2)) = arctan(5/4) ≈ 51.34 degrees
- Angle DAB: arctan((0-5)/(2-2)) = arctan(-5/0) = undefined

Since the angles are not all right angles (90 degrees), the figure ABCD is not a square.

Therefore, the figure ABCD is not a square.

Let us calculate the lengths of the sides of the quadrilateral ABCD:

$$AB=\sqrt{(-2-0)^2+(0-5)^2}=\sqrt{(-2)^2+(-5)^2}=\sqrt{4+25}=\sqrt{29}$$

$$BC=\sqrt{(0-2)^2+(5-0)^2}=\sqrt{(-2)^2+(5)^2}=\sqrt{4+25}=\sqrt{29}$$

$$CD=\sqrt{(2-0)^2+(0+5)^2}=\sqrt{(2)^2+(5)^2}=\sqrt{4+25}=\sqrt{29}$$

$$DA=\sqrt{(0+2)^2+(-5-0)^2}=\sqrt{(2)^2+(-5)^2}=\sqrt{4+25}=\sqrt{29}$$

Since all the sides of the quadrilateral ABCD are equal, it is a rhombus.

Now, let us calculate the diagonals of the quadrilateral ABCD:

$$AC=\sqrt{(-2-2)^2+(0+0)^2}=\sqrt{(-4)^2+(0)^2}=\sqrt{16+0}=\sqrt{16}=4$$

$$BD=\sqrt{(0-0)^2+(5+5)^2}=\sqrt{(0)^2+(10)^2}=\sqrt{0+100}=\sqrt{100}=10$$

Since the diagonals are not equal, i.e., AC≠BD, the quadrilateral ABCD is not a square.