Arbitrary Points

If you partition the diagram as follows:

then $A \cong A'$ , $B \cong B'$ , $C \cong C'$ , and $D \cong D'$ , so that the shaded part equals the unshaded part, and $50\% = \boxed{0.5}$ of the square is shaded.

I m shocked that no one did as I did so I'm sharing my method too(lol it's a lazy one)

Hence $\dfrac{1}{2} or \boxed{0.5}$ of the sq is shaded.

Denote the side of the square as x and the height of the lower triangle as y.

Shaded area : Area of square
= Lower triangle's area + Upper triangle's area : Area of square
= [ (1/2) × (lower base) × y ] + [ (1/2) × (upper base) × (x – y) ] : base × height
= 0.5xy + 0.5x(x – y) : x²
= 0.5x[ y + (x – y) ] : x²
= 0.5x² : x²
= 1 : 2

The problem is worded that whatever point we chose the result will be the same. Take the case where the arbitrary point lies in the middle of the square. Evidently is the answer: 0.5