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]
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°
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