State and Prove Alternate Segment Theorem on circles.

PLACEHOLDER ANSWER - this is a proof question. The theorem states that the angle between a tangent and a chord through the point of contact equals the angle subtended by the chord in the alternate segment. For a tangent at $$A$$ and chord $$AB$$, draw the diameter $$AD$$; then $$\angle DBA = 90^\circ$$ and the tangent is perpendicular to $$AD$$. So the tangent-chord angle equals $$90^\circ - \angle DAB = \angle ADB$$, and $$\angle ADB$$ equals the angle $$ACB$$ at any point $$C$$ of the alternate segment (same segment).