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    question on properties of circle 3

    Referring to the attached pic:
    (1) prove triangle BTQ ~ triangle QTA
    (2) find TQ
    (3) find the values of the radii
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  2. #2
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    Hello ukorov
    Quote Originally Posted by ukorov View Post
    Referring to the attached pic:
    (1) prove triangle BTQ ~ triangle QTA
    (2) find TQ
    (3) find the values of the radii
    Produce $\displaystyle QO_2$ to meet the circle at $\displaystyle R$. Then $\displaystyle \angle TBQ = \angle BRQ$ (alternate segment)

    $\displaystyle \Rightarrow \angle TBQ = \tfrac12 \angle BO_2Q$ (angle at centre)

    $\displaystyle \angle TAO_1 = \angle TBO_2 = 90^o$ (angle between tangent and radius)

    $\displaystyle \Rightarrow AO_1 \parallel BO_2$ (corresponding angles equal)

    $\displaystyle \Rightarrow \angle AO_1T = \angle BO_2Q$ (corresponding angles)

    But $\displaystyle \angle AQT =\tfrac12\angle AO_1T$ (angle at centre)

    $\displaystyle \Rightarrow \angle AQT = \tfrac12\angle BO_2Q =\angle TBQ$ (proved)

    So in $\displaystyle \triangle$'s $\displaystyle BTQ, QTA$:
    $\displaystyle \angle AQT = \angle TBQ$ (proved)

    $\displaystyle \angle ATQ = \angle QTB$ (same angle)
    $\displaystyle \Rightarrow \triangle BTQ \sim \triangle QTA$

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello ukorovProduce....
    alright that makes sense now.
    for question (2), is it to first prove that triangle AQB is right-angled, use Pythagarus Th. to find AB, and the rules of similar triangle to find TA, TO1, QO1, QO2?
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    Hello ukorov
    Quote Originally Posted by ukorov View Post
    alright that makes sense now.
    for question (2), is it to first prove that triangle AQB is right-angled, use Pythagarus Th. to find AB, and the rules of similar triangle to find TA, TO1, QO1, QO2?
    Yes. Note that $\displaystyle PA=PQ=PB$ (tangents from point to circle). So $\displaystyle Q$ lies on a circle that has $\displaystyle AB$ as a diameter. So $\displaystyle \angle AQB =$ ?

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello ukorovYes. Note that $\displaystyle PA=PQ=PB$ (tangents from point to circle). So $\displaystyle Q$ lies on a circle that has $\displaystyle AB$ as a diameter. So $\displaystyle \angle AQB =$ ?

    Grandad
    i saw that but i used angle AQT = a and angle BQR = (180 - a)/2 instead. with adj. angle on straight line, angle AQB = 90
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