Exercise 30 Page 769

We want to prove that AB is tangent to ⊙ O given that AB⊥ OP at P.

We will prove it by contradiction using an indirect proof. To prove a statement by contradiction, we begin by assuming that the conclusion is false. In this case, we will assume that AB is not tangent to ⊙ O. With this, we have two cases.

• Case I: AB does not intersect ⊙ O.
• Case II: AB intersects ⊙ O at two points.

  Let's examine each case separately.

  Case I

  If AB does not intersect ⊙ O, then point P is not on ⊙ O.

  

  However, we know from the given diagram, P lies on the circle and OP is a radius. Therefore, the result of Case I contradicts this.

  Case II

  Now, let's consider the case that AB intersects ⊙ O at two points, P and Q. Remember, it is given that AB⊥ OP at P.

  

  In this case, both OP and OQ are radii of ⊙ O. Since the radii of a circle are congruent, we have an isosceles triangle △ POQ. By the Isosceles Triangle Theorem, ∠ PQO is also a right angle.

  

  However, this contradicts that the sum of the interior angles of a triangle is 180^(∘). A triangle cannot have two interior angles measuring 90^(∘).

  Conclusion

  Because both cases resulted in contradictions, we can conclude that AB intersects ⊙ O at only one point P given that AB⊥ OP at P. Therefore, it must be tangent to ⊙ O.