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Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the center.
Let us consider a circle with center O and PA and PB are two tangents to the circle from an external point P
To Prove : Angle between two tangents is supplementary to the angle subtended by the line segments joining the points of contact at center, i.e. ∠APB + ∠AOB = 180°
As AP and BP are tangents to given circle,
OA ⏊ AP and OB ⏊ BP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP, By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠AOB + ∠APB = 360°
90° + 90° + ∠AOB + ∠APB = 360°
∠AOB + ∠APB = 180°