Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: AB is a tangent at point P to the circle.

To prove: OP ⊥ AB

Construction: Take a point Q on AB other than P and join OQ

Proof: If point Q lies inside the circle, then AB will become a secant and not a tangent to the circle

∴ OQ > OP

This happens with every pint on the line AB except the point P.

OP is the shortest of all the distances of the point O to the points of XY

∴ OP ⊥ AB, because shortest side is the perpendicular.