now use de similar to show the iii)
is very hard write in latex in this forum
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
originally posted by xwrathbringerx
Clues for you
angle BQP =90 deg halfarc of diameter
angle OQT =90 drg tangent radius property
angle x = half arc BQ tangent chord property
angle BPQ = half arc BQ angle between chords at circle property
angle BPQ = x
use the property of angle plus complement = 90 deg to define additional angles