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20/06/2021 11:09 am
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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 segment joining the points of contact at the center.
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20/06/2021 11:11 am
From the above diagram, it is seen that the line segments OA and PA are perpendicular.
So, ∠OAP = 90°
In a similar way, the line segments OB ⊥ PB and so, ∠OBP = 90°
Now, in the quadrilateral OAPB,
∴∠APB+∠OAP +∠PBO +∠BOA = 360° (since the sum of all interior angles will be 360°)
By putting the values we get,
∠APB + 180° + ∠BOA = 360°
So, ∠APB + ∠BOA = 180°