1. ## Circle Geometry

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

2. $\displaystyle \measuredangle OQB = 2x \Rightarrow \measuredangle QOP = 180 - 2x{\text{ and }}\vartriangle OTQ \cong \vartriangle OTP \Rightarrow$

$\displaystyle \begin{gathered} \measuredangle TOC = \measuredangle TOP = 90 - x \hfill \\ \Rightarrow \measuredangle OTC = \measuredangle OTP = x \Leftrightarrow \measuredangle PTQ = 2x \hfill \\ \end{gathered}$

$\displaystyle {\text{The }}\vartriangle QOB{\text{ have 2 sizes same}}{\text{, because }}\overline {BQ} \wedge \overline {BO} {\text{ are radius}}$
$\displaystyle \begin{gathered} \Rightarrow \measuredangle OBQ = 90 - x, \hfill \\ {\text{look that }}\measuredangle BQP = 90 \hfill \\ \end{gathered}$

$\displaystyle {\text{because }}\overline {BP} {\text{ is diameter and how I prove the angles of }}\vartriangle OTQ$

$\displaystyle \Rightarrow \vartriangle TOQ \sim \vartriangle PBQ$

now use de similar to show the iii)

is very hard write in latex in this forum

3. ## circle geometry

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

bj