Answer:
Draw center O. Since PCQ is tangent to the circle, it is known that OC is perpendicular to PQ; that is, <OCQ = 90. Since <OCQ = 90 and <BCQ = x, <OCB = 90 - x. Since O is the center and B, C lie on the circle, OC = OB. By definition, then, triangle OCB is isosceles. Since OCB is isosceles, <OBC = <OCB = 90 - x. Since the sum of the internal angles of a triangle is 180, <OCB + <OBC + <BOC = 180, that is, (90 - x) + (90 - x) + <BOC = 180. From simple algebra it follows that <BOC = 2x.
Since A also lies on the circle, OA = OB = OC, and in fact, since AB = BC (given), triangles OBC and OAB are congruent by SSS. Since they are congruent, it follows that <BOC = <AOB = 2x.
Then, <AOC = <AOB + <BOC = 2x + 2x = 4x. Since <AOC = 4x, by the Inscribed Angle Theorem, <ADC = <AOC / 2 = 2x.
And hence, <ADC = 2x (in degrees).
Explanation: