Answer :

Let us consider a circle with center O, OP be a radius and XY be a tangent at point P.

To prove: OP ⊥ XY

Proof: Let’s take a point Q on XY other than P.

Clearly, point Q lies outside the circle because if Q lies inside the circle then XY will be a secant (as it will intersect the circle at two points)

Point P is on circle and point Q is outside the circle

⇒ OP < OQ

This is true for all points on XY other than P

⇒ OP is the shortest distance between point P and line XY.

⇒ OP ⊥ XY [Shortest side is perpendicular]

Hence, proved!

**OR**

Let us consider a circle with center O. PQ and PR be two tangents from an external point P to the circle.

To prove: ∠QPR + ∠QOR = 180°

Proof: In quadrilateral PROQ, we have

∠QPR + ∠PQO + ∠QOR + ∠ORP = 360°

Now, ∠OQP = ∠ORP = 90° [Tangent at a point on a circle is perpendicular to the radius through point of contact]

⇒∠QPR + 90° + ∠QOR + 90° = 360°

⇒∠QPR + ∠QOR = 180°

Hence, proved!

Rate this question :