Given: AB is a tangent at the point P of a circle with center O and OP is a radius through the point P. To prove: OP and AB are perpendicular to each other i.e. OP ⊥ AB.

Construction: Any other point Q is taken on the tangent AB, O, Q are joined.

Proof: Any other point on AB except P is outside the circle; ∴ OQ intersects the circle at a point. Let R be the point of intersection.

∴ OR < OQ [∴ R is a point between O, Q]

Again, OR = OP [∴ radii of the same circle]

∴ OP < OQ

∴ The point Q is any point on AB,

∴ OP is the least of all the line segments drawn from the center O to the tangent AB.

Again, the least distance is perpendicular distance.

∴ OP ⊥ AB (proved)