Question

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 centre.

Answer 1

Given, PA and PB are the tangents drawn from a point P to a circle with center O. Also, the line segments OA and OB are drawn.

To prove: ∠APB +  ∠AOB = 180°

We know that the tangent to a circle is perpendicular to the radius through the point of contact

∴ PA ⊥ OA

\(\Rightarrow\) ∠OAP = 90°

PB ⊥ OB

\(\Rightarrow\) ∠OBP = 90°

∴ ∠OAP + ∠OBP = (90° + 90°) = 180°   .......(i)

But we know that the sum of all the angles of a quadrilateral is 360°.

∴ ∠OAP + ∠OBP + ∠APB + ∠AOB = 360°  .....(ii)

From (i) and (ii), we get:

∠APB + ∠AOB = 180°