[b]Conjecture: [/b]If a line is tangent to a circle, then the line is perpendicular to a radius of the circle at the point of tangency.[br][br][b]Contrapositive of Conjecture: [/b]If a line that intersects a circle is not perpendicular to a radius of the circle at the point of intersection, then the line is not tangent (that is- the line intersects at more than one point of the circle). [br][br][b]Proof (By contrapositive): [/b]Construct the circle centered at point O. Now, consider the line [i]l[/i] that intersects circle O at the point P. By Euclid's Proposition 13, we know that the radius at point P forms two angles, call them [math]\alpha[/math] and [math]\beta[/math] such that [math]\alpha+\beta=180^\circ[/math]. [br][br]Assume that [math]\alpha\ne\beta\ne90^\circ[/math]. That is, assume [i]l[/i] is not perpendicular to a radius of the circle at P. Then, without loss of generality, we can also assume that [math]\beta>90^\circ>\alpha[/math]. [br][br]Now, form the isosceles triangle [math]\triangle OPQ[/math] such that Q lies on [i]l, [/i]the angle [math]\angle OPQ[/math] coincides with angle [math]\alpha[/math], and [math]\angle OPQ=\angle OQP=\alpha[/math]. By Euclid's Proposition 6, the segment [math]\overline{OQ}[/math] must equal the length of the radius. Thus, the point Q must be a point on the circle. Then, we know that [i]l [/i]intersects the circle at P and Q. Because [i]l [/i]intersects the circle at more than one point, [i]l[/i] is not tangent to the circle. [br][br]Therefore, the contrapositive of the conjecture above is true and thus the conjecture is also true.