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 to the centre.

Answer 1

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

To prove:$\angle$APB + $\angle$AOB = 180°
Proof:
We know that the tangent to a circle is perpendicular to the radius through the point of contact.
∴ PA ⊥ OA ⇒ $\angle$OAP = 90°
PB ⊥ OB ⇒ $\angle$OBP = 90°
∴ $\angle$OAP + $\angle$OBP = (90° + 90°)= 180° ......................(i)
But we know that the sum of all the angles of a quadrilateral is 360°.
∴ $\angle$OAP + $\angle$OBP + $\angle$APB + $\angle$AOB = 360° .................(ii)
From (i) and (ii), we get:
$\angle$APB + $\angle$AOB = 180°