Angle Formed By Chord and Tangent

Prove that the measure of an angle formed by a chord and a tangent is equal to half the measure of the intercepted arc.

Given circle O, tangent DA where point A is on circle O, and chord BA for circle O. To show that the measure of an angle formed by a chord and a tangent is equal to half the measure of the intercepted arc, I will use proof by cases as the first case has already been proved.

Case 1 is where chord BA is a diameter so point O is on chord BA such that B-O-A. This makes OA and OB radii of circle O so according to Theorem 6.7, tangent DA would be perpendicular to radii OA. This means that the angle measure on both sides is 90 which is indeed, as proved earlier (problem 6.3.2), half of the intercepted arc.

Case 2 is where chord BA is smaller than the diameter. AP1 allows us to construct a line parallel to the tangent DA going through point B, line BC. This creates angle ABC with the intercepted arc, AC. Theorem 6.9 states that the measure of this angle, angle ABC is equal to half of the measure of arc AC.

As lines DA and BC are parallel and secant of chord AB is a transversal, then angles DAB and ABC are alternating interior angles (Definition 4.7) and they are congruent (Theorem 4.27). Therefore, as ABC is half the measure of the intercepted arc, then angle DAB is also half the measure of the intercepted arc; or the measure of an angle formed by a chord and a tangent is equal to half the measure of the intercepted arc.