As given in the question, ABCD is a square. The angles between the sides of the square are always 90°. The diagonal divides this angle exactly into two halves.
Therefore, the angle BCA is
[tex]\angle BCA=\frac{90^0}{2}=45^0^{}_{}[/tex]
The total angle at the centre where the diagonals meet is 360°.
These diagonals cut it into four equal parts.
Therefore, angle AWD is
[tex]\angle AWD=\frac{360^0}{4}=90^0[/tex]
