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Thread: In a quadrilateral ABCD, AB=AC=AD

  1. #1
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    jharkhand
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    In a quadrilateral ABCD, AB=AC=AD

    Hence show that:

    <BAD = 2(<CBD + <CDB)
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  2. #2
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    Hello, snigdha!

    Here's one solution . . . (There must be many others.)


    In quadrilateral $\displaystyle ABCD\!:\;\;AB\,=\,AC\,=\,AD$

    Show that: .$\displaystyle \angle BAD \;=\; 2\left(\angle CBD + \angle CDB\right)$

    Quadrilateral $\displaystyle ABCD$ is a "kite".



    Code:
                A
                o
               /|\
              /θ|θ\
             /  |  \
            /   |   \
           /    |    \
          /     |     \
         / θ'   |   θ' \
      B o - - - + - - - o D
          * α   |   α *
            * α'|α' *
              * | *
                o
                C

    Let $\displaystyle \theta \,=\, \angle BAC \,=\, \angle DAC$

    Then $\displaystyle \theta' \,=\, \angle ABD = \angle ADB$ . where $\displaystyle \theta' \,=\,90^o-\theta$


    Let $\displaystyle \alpha \,=\, \angle CBD \,=\, \angle CDB$

    Then $\displaystyle \alpha' \,=\,\angle ACB \,=\,\angle ACD$ . where $\displaystyle \alpha' \,=\,90^o-\alpha$


    Since $\displaystyle \Delta ABC$ is isosceles: .$\displaystyle \angle ABC \,=\,\angle ACB$

    . . Hence: .$\displaystyle \theta' + \alpha \:=\:\alpha' \quad\Rightarrow\quad (90^o-\theta) + \alpha \:=\:(90^o-\alpha) $


    And we have: .$\displaystyle \theta \:=\:2\alpha \quad\Rightarrow\quad 2\theta \:=\:2(2\alpha)$


    Therefore: .$\displaystyle 2\theta \quad=\;\; 2\;\;(\;\;\alpha \quad+\quad\;\; \alpha\;\;)$
    . . . . . . . . .$\displaystyle \downarrow\qquad\qquad\quad\, \downarrow \qquad\qquad \downarrow$

    . . . . . . $\displaystyle \angle BAD \;=\;2(\angle CBD + \angle CDB)$

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