Let ABCD be a quadrilateral with center o.
The angle BAo,DAO is equal as AB and AD are tanglents
let each angle equal = a;
The angles at B are similarly equal to each other. Let each of them equal b.
Similarly for vertices C and D.
The sum of the angles at the centre is 360 deg.
The sum of the angles of ABCD is 360 deg.
2(a + b + c + d) = 360
a + b + c + d = 180.
From triangle AOB, angle BOA = 180 - (a + b).
From triangle COD, angle COD = 180 - (c + d).
Angle BOA + angle COD = 360 - (a + b + c + d)
= 360 - 180
= 180 deg.
Thus AB and CD subtend supplementary angles at O.