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Thread: Cyclic quadrilateral

  1. #1
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    Cyclic quadrilateral

    1. ABCP are points on the circle.ABC is a isocelels triangle with AB=AC and angle bisector of B meets circle at P. If AP and BC is extended to meet at Q, show that AC=QC.
    Please let me know the steps to solove the above problem
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \widehat{ABC}=\widehat{ACB}$

    $\displaystyle \widehat{ABP}=\widehat{PBC}=\widehat{PAC}=\frac{1} {2}\widehat{ACB}$

    $\displaystyle \widehat{ACB}=\widehat{PAC}+\widehat{AQC}=\frac{1} {2}\widehat{ACB}+\widehat{AQC}\Rightarrow$

    $\displaystyle \Rightarrow\widehat{AQC}=\frac{1}{2}\widehat{ACB}= \widehat{PAC}\Rightarrow\Delta ACQ$ isosceles $\displaystyle \Rightarrow AC=CQ$
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  3. #3
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    Hello, reshma!

    1. $\displaystyle ABCP$ are points on the circle.
    $\displaystyle ABC$ is an isocelels triangle with $\displaystyle AB=AC$, and angle bisector of $\displaystyle B$ meets circle at $\displaystyle P.$
    If $\displaystyle AP$ and $\displaystyle BC$ is extended to meet at $\displaystyle Q$, show that: $\displaystyle AC=QC.$

    Since $\displaystyle \angle ABC = \angle ACB,\;\text{arc}(AC) = \text{arc}(AB)$

    Since $\displaystyle \angle ABP = \angle PBC,\;\text{arc}(AP) = \text{arc}(PC)$

    Let $\displaystyle \text{arc}(AP) \,=\,\text{arc}(PC) \,=\,a$
    . . Then: .$\displaystyle \text{arc}(AB) \,=\,2a$


    $\displaystyle \angle PAC$ is an inscribed angle.
    . . It is measured by: .$\displaystyle \tfrac{1}{2}\text{arc}(PC) \,=\,\tfrac{1}{2}a$

    $\displaystyle \angle AQB$ is formed by two secants.
    . . It is measured by: $\displaystyle \tfrac{1}{2}\left[\text{arc}(AB) - \text{arc}(PC)\right] \;=\;\tfrac{1}{2}\left[2a-a\right] \;=\;\tfrac{1}{2}a$

    Hence: .$\displaystyle \angle PAC \,=\,\angle AQB$, and $\displaystyle \Delta ACQ$ is isosceles.

    Therefore: .$\displaystyle AC \,=\,CQ$

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  4. #4
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    doubt

    Hi,

    Can u pls explain me the below part:

    is formed by two secants
    It is measured by:

    Thanks in advance
    Reshma
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