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Math Help - 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 ABCD\!:\;\;AB\,=\,AC\,=\,AD

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

    Quadrilateral ABCD is a "kite".



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

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

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


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

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


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

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


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


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

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

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