From an external point T, tangents are drawn to a circle with centre O, touching the circle at P and Q. Angle PTQ is acute. The diameter PB produced meets the tangent TQ at A. Let x=angle AQB.

(i) Prove that angle PTQ=2x

(ii) Prove that triangles PBQ and TOQ are similar.

(iii) Hence show that BQ*OT=2(OP)^2

Please help. I'm clueless.

Thanx in advance