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Thread: Triangle Question

  1. #1
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    Triangle Question

    In a triangle ABC where <A=120 degrees, Q and P are points on BC such that BP=BA and CQ=CA, where Q lies between B and P. Prove that <PAQ=30 degrees.

    I really could use a lot of help on this one.
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  2. #2
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    Hello, TheNcredibleHulk!

    Did you make a sketch?


    In triangle $\displaystyle ABC$ where $\displaystyle \angle A=120^o,\;P$ and $\displaystyle Q$ are points on $\displaystyle BC$
    such that: $\displaystyle BP=BA$ and $\displaystyle CQ=CA$, where $\displaystyle Q$ lies between $\displaystyle B$ and $\displaystyle P.$
    Prove that: .$\displaystyle \angle PAQ\,=\,30^o$
    Code:
        B o
            * *
              *   *
                *  θ  *
                  *       *    Q
                    *         o    P
                      *           o
                        *       *     *
                          *   *           *
                            o  *  *  *  *  *  o C
                           A

    Draw line segment $\displaystyle AQ$.

    Let $\displaystyle \angle B \,=\,\theta$ . . . Then $\displaystyle \angle C \,=\,60-\theta$


    In isosceles $\displaystyle \Delta ABP\!:\;\angle BPA = \angle BAP$

    . . $\displaystyle \theta + \angle BPA + \angle BAP \:=\:180^o \quad\Rightarrow\quad \angle BPA \:=\:90 - \tfrac{1}{2}\theta \:=\:\angle QPA$


    In isosceles $\displaystyle \Delta CAQ\!:\;\angle CQA = \angle CAQ$

    . . $\displaystyle (60-\theta) + \angle CQA + \angle CAQ \:=\:180^o \quad\Rightarrow\quad \angle CQA \:=\:60 + \tfrac{1}{2}\theta \:=\:\angle PQA$



    In $\displaystyle \Delta APQ\!:\;\angle PAQ + \angle PQA + \angle QPA \:=\:180^o$


    Therefore: .$\displaystyle \angle PAQ + \left(90^o-\tfrac{1}{2}\theta\right) + \left(60^o +\tfrac{1}{2}\theta\right) \:=\:180^o \quad\Rightarrow\quad \angle PAQ \:=\:30^o $

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