Opposite Angles of Cyclic Quad

What you say about angles at C and D ? 

You may drag all the points on the circumfrence

Can you recollect any theorem ? 

∠ACB = \frac{1}{2}∠BOA      

angle at the centre thoerem

∠BDA = \frac{1}{2}∠AOB

∠C + ∠D = \frac{1}{2}[∠BOA +∠AOB]

∠C +∠D = \frac{1}{2}× 360° = 180°

Opposite angles of a cyclic quad are supplementary

O = (5.66, -1.12)
H = (0.96, -3.1)
c = Circle(O, H)
A = Point(c)
B = Point(c)
e = Arc(c, B, A)
d = Arc(c, A, B)
h = Segment(B, O)
i = Segment(O, A)
α = Angle(A, O, B)
C = Point(e)
n = Segment(A, C)
p = Segment(C, B)
ε = Angle(A, C, B)
D = Point(d)
j = Segment(A, D)
k = Segment(D, B)
γ = Angle(B, D, A)
δ = Angle(B, O, A)
DragMe = 4 (slider from 0 to 4, step 1)