Very nice math problem. Anyone care to tackle?

You just need to know that the angle external to a circle determined by a secant line and a tangent line, or by a pair of tangent lines is 1/2 the difference of the far and near arcs. You can use the same theorem for the angle α which is inside the circle. In this case, the "near arc" has measure 0°.

So

2α = arc(BD)

2ß = arc(AD) - arc(BC), and

2γ = arc(AB) - arc(CD).

Summing gives

2(α + ß + γ) = arc(BD) + arc(AD) + arc(AB) - arc(BC) - arc(CD).

Now, arc(BD) + arc(AD) + arc(AB) = 360°---this is the full circle. Moreover, arc(BC) + arc(CD) = arc(BD) = 2α.

Dividing out 2 and simplifying.

α + ß + γ = 180° - α ==> ß + γ = 180° - 2α

as required.