Question

In following figure from exterior point ‘P’ two tangent line PA and PB touches the circle at A and B, then prove that PAOB is a cyclic quadrilateral where O is the  center of circle.

Answer 1

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

OA ⊥ PA and OB ⊥ PB

∴ ∠OAP = ∠OBP = 90°

⇒ ∠OAP + ∠OBP = 90° + 90° = 180°

In quadrilateral PAOB

∠PAO + ∠AOB + ∠OBP + ∠BPA = 360°

⇒ ∠AOB + ∠BPA = 360° – 180°

⇒ ∠AOB + ∠BPA=180°

∵ Sum of opposite angle of quadrilateral is = 180°.

∴ Quadrilateral PAOB is a cyclic quadrilateral