Q. 104.4( 345 Votes )
To Prove: ∠ APB + ∠ BOA = 180°
Let us consider a circle centred at point O.
Let P be an external point from which two tangents PA and PB are drawn to the circle which touches the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends
∠ AOB at centre O of the circle.
It can be observed that
OA ⊥ PA (radius of circle is always perpendicular to tangent)
Therefore, ∠ OAP = 90°
Similarly, OB ⊥ PB
∠ OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360°
∠ OAP +∠ APB+∠ PBO +∠ BOA = 360°
90° + ∠ APB + 90° + ∠ BOA = 360°
∠ APB + ∠ BOA = 180°
Hence, it can be observed that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
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