(1) Prove that Tangent to a circle at a point is perpendicular to the radius through the point of contact.
Given: A circle C(O,r) and a tangent AB at a point P.
To Prove: OP⊥AB
Construction: Take any Point Q, other than P, on the tangent AB. Join OQ. Suppose OQ meets the circle at R.Proof: We know that among all line segments joining the point O to a point on AB the shortest one is perpendicular to AB. So, to prove that OP⊥AB, it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.
Cleraly,      OP=OR               [Radii of the same circle]
Now,          OQ=OR+RQ
⇒       OQ>OR
⇒       OQ>OP        [OP=OR]
⇒       OQ<OQ
Thus, OP is shorter than any other segment joining O to any point of AB.
Hence, OP⊥AB.

(2) Prove that from a point, lying outside a circle, two and only two tangents can be drawn to it.
When the point lies outsides the circle, there are exactly two tangents to circle from a point which lies outside the circle. As shown in figure.

(3) Prove that the lengths of the two tangents drawn from an external point to a circle are equal.
Given: AP and AQ are two tangents from a point A to a circle C(O,r).
To Prove: AP=AQ
Construction: Join OP,OQ and OA
Proof: In order to prove that AP=AQ we shall first prove that ΔOPA≅ΔOQA.
Since a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OP⊥AB and OQ⊥AQ
⇒∠OPA=∠OQA=90∘….(i)
Now, in right triangle OPA and OQA, we have
OP=OQ                        [Radii of a circle]
∠OPA=∠OQA             [From (i)]
And,             OA=OA                        [Common]
So, by RHS-Criterion of congruence, we get
ΔOPA≅ΔOQA
⇒         AP=AQ