Let C be a center of a circle, and let AB be the diameter that crosses point C:
Point D is a point that lies on the circumference, which means line CD is a radius. Lets call ∠BAD α, and ∠ABD β:
Since CB = CD, then ∠CBD = ∠CDB. Also, since CA = CD, then ∠CAD = ∠CDA:
Since angles in a triangle add to 180°:
This proves that ∠ADB is 90°.