Given A circle with center 'O'
Two tangents PQ and PT from an external point P.
Let Q, T be the points of contact
R.T.P ∠P and ∠QOT are supplementary
Proof OQ ⟂PQ
[radius is perpendicular to the tangent at the point of contact] also OT ⟂PT
∠OQP + ∠OTP = °+° = °
∠OTP+∠TPQ+∠PQO+∠QOT = 360°(angle sum property)
° + ∠P+∠QOT = 360°
∠P+∠QOT = 360° - ° = ° Hence proved