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Math Help - Circle Geometry

  1. #1
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    Exclamation 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

    Please help. I'm clueless.

    Thanx in advance
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  2. #2
    Member Nacho's Avatar
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    <br />
\measuredangle OQB = 2x \Rightarrow \measuredangle QOP = 180 - 2x{\text{ and }}\vartriangle OTQ \cong \vartriangle OTP \Rightarrow <br />

    <br />
\begin{gathered}<br />
  \measuredangle TOC = \measuredangle TOP = 90 - x \hfill \\<br />
   \Rightarrow \measuredangle OTC = \measuredangle OTP = x \Leftrightarrow \measuredangle PTQ = 2x \hfill \\ <br />
\end{gathered} <br />

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

    <br />
{\text{because }}\overline {BP} {\text{ is diameter and how I prove the angles of }}\vartriangle OTQ<br />

    <br />
 \Rightarrow \vartriangle TOQ \sim \vartriangle PBQ<br />

    now use de similar to show the iii)

    is very hard write in latex in this forum
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  3. #3
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    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
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