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Thread: circle geometry help 2

  1. #1
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    circle geometry help 2

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  2. #2
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    Hello andy69

    The question is shown in the diagram I have attached.

    Proof

    (i) $\displaystyle \angle QXB + \angle BXA + \angle AXP = 180^o$ (angles on a straight line)


    $\displaystyle \angle BXA = 90^o$ (given)


    $\displaystyle \therefore \angle QXB + \angle AXP = 90^o$


    But $\displaystyle \angle PAX + \angle AXP = 90^o$ (angle sum of $\displaystyle \triangle APX$)


    $\displaystyle \therefore \angle QXB = \angle PAX$
    $\displaystyle =\angle QBX$ (angles in same segment)
    (ii) $\displaystyle BQ = XQ$ (isosceles $\displaystyle \triangle BQX$, from part (i))

    $\displaystyle \angle BXC = 90^o$ (given)


    $\displaystyle \therefore X$ lies on a circle with diameter BC (angle in semicircle)


    But $\displaystyle BQ = XQ$ (proven)


    $\displaystyle \therefore Q$ is the centre of the circle through $\displaystyle B, \;X$ and $\displaystyle C$


    $\displaystyle \therefore BQ = QC$


    Grandad
    Attached Thumbnails Attached Thumbnails circle geometry help 2-untitled.jpg  
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello andy69

    The question is shown in the diagram I have attached.

    Proof

    (i) $\displaystyle \angle QXB + \angle BXA + \angle AXP = 180^o$ (angles on a straight line)


    $\displaystyle \angle BXA = 90^o$ (given)


    $\displaystyle \therefore \angle QXB + \angle AXP = 90^o$


    But $\displaystyle \angle PAX + \angle AXP = 90^o$ (angle sum of $\displaystyle \triangle APX$)


    $\displaystyle \therefore \angle QXB = \angle PAX$
    $\displaystyle =\angle QBX$ (angles in same segment)
    (ii) $\displaystyle BQ = XQ$ (isosceles $\displaystyle \triangle BQX$, from part (i))

    $\displaystyle \angle BXC = 90^o$ (given)


    $\displaystyle \therefore X$ lies on a circle with diameter BC (angle in semicircle)


    But $\displaystyle BQ = XQ$ (proven)


    $\displaystyle \therefore Q$ is the centre of the circle through $\displaystyle B, \;X$ and $\displaystyle C$


    $\displaystyle \therefore BQ = QC$


    Grandad
    hey thanks for the help you mind me asking how do you attach the pictures?
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  4. #4
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    Hello andy69
    Quote Originally Posted by andy69 View Post
    hey thanks for the help you mind me asking how do you attach the pictures?
    I have a program called MWSnap that enables me to capture any part of the screen, to clipboard. Paste it into Paint; crop as necessary; save as .jpg. Quick and simple. Another example attached!

    Grandad
    Attached Thumbnails Attached Thumbnails circle geometry help 2-untitled.jpg  
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