Step-by-step explanation:
The various theorems involved include ...
- angle sum theorem (angles in a triangle total 180°)
- alternate interior angles theorem (said angles are congruent)
- inscribed angle theorem (and "alternate")
- angle addition postulate
For the purpose here, we wll use the notation D.1 to refer to the angle labeled "1" at vertex D.
The argument goes something like the following.
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∠D.1 +∠R +∠S.1 = 180° . . . . angle sum theorem
∠S.1 +∠S.2 +∠S.3 +∠S.4 = 180° . . . . angle addition postulate (∠RSW is a straight angle, equal to 180°)
∠D.1 +∠R +∠S.1 = ∠S.1 +∠S.2 +∠S.3 +∠S.4 . . . . substitution property
∠D.1 = ∠S.2 . . . . alternate interior angles theorem
∠S.2 +∠R +∠S.1 = ∠S.1 +∠S.2 +∠S.3 +∠S.4 . . . . substitute for ∠D.1
∠R = ∠S.3 +∠S.4 . . . . subtraction property (subtract ∠S.1 +∠S.2)
∠R = ∠S.3 +∠S.2 . . . . substitute ∠S.2 for equal ∠S.4 (given)
∠R = arc DS/2 . . . . inscribed angle is half the arc measure
∠TSD = ∠S.2 +∠S.3 = arc DS/2 . . . . substitution, angle addition, angle naming
TS is tangent at S . . . . converse of "inscribed angle alternate" where the vertex of an inscribed angle is a point of tangency
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Additional comment
The alternate to the inscribed angle theorem says the angle is half the measure of the intercepted arc even as the vertex of the inscribed angle becomes a point of tangency (one ray of the angle is a tangent, so a zero-length chord). The converse of this says the vertex is a point of tangency if the intercepted arc is twice the measure of the angle, as in this problem.