If two diameters of a circle intersect each other at right angles, then the quadrilateral formed by joining here end points is a :

From given figure , we have
In quadrilateral ABCD , we can see that
Diagonal AB = Diagonal CD ......... ( ⅰ )
AO = OB ......... ( ⅱ )
DO = CO ......... ( ⅲ )
∠DOB = 90° ......... ( ⅳ )
From ( ⅰ ) , ( ⅱ ) , ( ⅲ ) and ( ⅳ ) , we get
From above , it is clear that In quadrilateral ABCD , diagonals are equal and they bisect each other at 90° .
⇒ Quadrilateral ABCD is square.