Answer:
Given, a circle with centre O and a tangent AB at point P on circle. To prove: \[OP\bot AB\]. Construction: Take another point Q on AB and join OQ. Proof: Since Q is a point on AB (other than P) \[\therefore \] Q lies outside the circle. Let OQ intersect the circle at R, Then, \[OR<OQ\] ?(i) But \[OP=OR\] (radii of circle) ?(iii) \[\therefore OP<OQ\] (from (i) and (ii)) Thus, OP is shorter than any other line segment joining O to any point on AB. But the shortest distance between a point and a line is the perpendicular distance. \[\therefore \,\,\,OP\bot AB\] Hence Proved.
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