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 pointsof contact to the centre.
Given: PA and PB are the tangent drawn from a point P to a circle with centre O. Also, the line segments OA and OB are drawn.
To Prove: ∠APB + ∠AOB = 180°
Proof: We know that the tangent to a circle is perpendicular to the radius through the point of contact.
∴ PA ⊥OA ⇒ ∠OAP = 90°, and
PB ⊥ OB ⇒ ∠OBP = 90°.
∴ ∠OAP + ∠OBP = 90°.
Hence, ∠APB + ∠AOB = 180°
[∵ sum of the all the angles of a quadrilateral is 360°]
Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching 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 center O of the circle.
It can be observed that
OA (radius) ⊥ PA (tangent)
Therefore, ∠OAP = 90°
Similarly, OB (radius) ⊥ PB (tangent)
∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB+∠PBO +∠BOA = 360º
90º + ∠APB + 90º + ∠BOA = 360º
∠APB + ∠BOA = 180º
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.