Given: AB is a tangent to a circle with center O, touching the circle at point P. OP is the radius drawn to the point of contact P. To prove: OP ⊥ AB, i.e., the radius OP is perpendicular to the tangent AB. Construction: Take any other point Q on the tangent AB. Join points O and Q. Proof: Any point on the tangent AB other than the point of contact P lies outside the circle. Therefore, the line OQ will intersect the circle at a point. Let that point of intersection be R. ∴ OR < OQ [∵ R lies between O and Q] Also, OP < OQ Since Q is any point on the tangent AB, the line segment OP from the center O to the tangent AB is the shortest among all such segments. And the shortest distance from a point to a line is the perpendicular distance. ∴ OP ⊥ AB (Proved)