Q8 of 12 Page 164

In the figure, O is the centre of the circle. Tangents PA and PB drawn from an external point P touch the circle at A and B respectively. Prove that PAOB is a cyclic quadrilateral.

Since tangents to a circle is perpendicular to the radius.

∴ OA ⊥ AP and OB ⊥ BP.


⟹ ∠OAP = 90° and ∠OBP = 90°


⟹ ∠OAP + ∠OBP = 90° + 90° = 180°........... (1)


In quadrilateral OAPB,


∠OAP + ∠APB + ∠AOB + ∠OBP = 360°


⟹ (∠APB + ∠AOB) + (∠OAP + ∠OBP) = 360°


⟹ ∠APB + ∠AOB + 180° = 360° [From (1)]


⟹ ∠APB + ∠AOB = 180°................ (2)


Thus from equations (1) and (2) it can be concluded that the quadrilateral PAOB is a cyclic quadrilateral.


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