Given: AB is a tangent to a circle centered at O, touching the circle at point P. OP is the radius drawn to the point of contact. To prove: OP ⊥ 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 some 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, and OP is the shortest distance from the center O to the tangent, and the shortest distance from a point to a line is the perpendicular distance, ∴ OP ⊥ AB (Proved)