If from an extern

Construction: Draw a circle (center O) with the given conditions i.e. external point P and two tangents PQ and PR.

To Prove: 2PQ = PO

We know that the radius is perpendicular to the tangent at the point of contact.

⇒ ∠OQP = 90°

We know that the tangents drawn to a circle from an external point are equally inclined to the segment, joining to the centre to that point.

⇒ ∠QPO = 60°

Consider ΔQPO,

Cos 60° = PQ/PO

⇒ � = PQ/PO

⇒ 2PQ = PO

Ans. Hence, proved that 2PQ = PO.