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Thread: circles

  1. #1
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    Unhappy circles

    Hello everyone !
    please give me an answer to the given circle geometry questions as fast as possible and if possible please today.

    Q1. Triangle ABC is inscribed in a circle. the bisector of angle BAC meets line segment BC at P and the circle at Q. line segment QC is joined. If angle BCQ = 38 degree , what is angle BAC ?

    Q2. AbCD is a cyclic trapezium in which AD is parallel to BC. Show that angle B = angle C ?
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  2. #2
    Super Member

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    Hello, Aveek Agrawal!

    1) Triangle $\displaystyle ABC$ is inscribed in a circle.
    The bisector of $\displaystyle \angle BAC$ meets side $\displaystyle BC$ at $\displaystyle P$ and the circle at $\displaystyle Q.$
    Line segment QC is drawn.
    If $\displaystyle \angle BCQ = 38^o$, find $\displaystyle \angle BAC.$
    Code:
                    B
                  * * *
              *    / \    *
            *     /   \     *
           *     /     \     * Q
          *     /       \     * 
          *    /         \ *  *
          *   /        *  \   *
          *  / θ   *       \  *
          * /  * θ          \ *
         A *- - - - - - - - -* C
            *               *
              *           *
                  * * *
    $\displaystyle AQ$ bisects $\displaystyle \angle A\!:\;\;\angle BAQ = \angle QAC = \theta$

    Draw chord $\displaystyle QC.$


    An inscribed angle is measure by one-half its intercepted arc.

    Since $\displaystyle \angle BCQ = 38^o$, then $\displaystyle \text{arc}\,BQ = 76^o$

    Since $\displaystyle \angle BAQ \;\;^m_=\;\;\tfrac{1}{2}\text{arc}\,BQ,\;\;\angle BAQ = 38^o$ . . . then $\displaystyle \angle QAC = 38^o$

    Therefore: .$\displaystyle \angle BAC \:=\:76^o$





    2) $\displaystyle ABCD$ is a cyclic trapezium in which $\displaystyle AD \parallel BC.$
    Show that: .$\displaystyle \angle B \:=\: \angle C$
    Code:
             A    * * *    D
              * - - - - - *
            */             \*
           */               \*
           /                 \
        B * - - - - - - - - - * C
          *                   *
          *                   *
    
           *                 *
            *               *
              *           *
                  * * *
    Since $\displaystyle AD \parallel BC$, then: /$\displaystyle \text{arc}\,AB \,=\,\text{arc}\,DC $
    . .
    Parallel chords intercept equal arcs.

    Add $\displaystyle \text{arc}\,AD$ to both sides:
    . . $\displaystyle \text{arc}\,AB + \text{arc}\,AD \;=\;\text{arc}\,DC + \text{arc}\,AD \quad\Rightarrow\quad \text{arc}\,BAD \:=\:\text{arc}\,ADC$


    Since: .$\displaystyle \begin{array}{ccc}\angle B & ^m_= & \tfrac{1}{2}\text{arc}\,ADC \\ \\[-4mm] \angle C & ^m_= & \tfrac{1}{2}\text{arc}\,BAD \end{array}\quad \text{then: }\angle B \:=\:\angle C$

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