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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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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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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°

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