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Math Help - Finding an angle

  1. #1
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    Finding an angle

    Triangle ABC where AB=AC. D is on AC such that BC=BD, and E is on AB where BE=ED=AD.

    What is angle EAD?
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  2. #2
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    is the answer 45 degrees
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  3. #3
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    Hello, Denny!

    Triangle ABC where AB=AC.
    D is on AC such that BC=BD.
    E is on AB such that BE=ED=AD.

    What is angle A?
    Code:
                      A
                      o
                     / \
                    /   \
                   /     \ D
                  /       o
                 /   *  *  \
              E o     *     \
               /    *        \
              /   *           \
             /  *              \
            / *                 \
           /*                    \
        B o - - - - - - - - - - - o C
    Let: \alpha \:=\:\angle A

    \Delta ABC is isosceles: . \angle B = \angle C
    Then: . \angle A + \angle B + \angle C \:=\:180^o \quad\Rightarrow\quad \alpha + 2\angle C \:=\:180^o \quad\Rightarrow\quad \angle C\:=\:\frac{180^o-\alpha}{2}

    Since BC = BD,\:\Delta BCD is isosceles: . \angle BDC \:=\:\angle C \quad\Rightarrow\quad \angle BDC \:=\:\frac{180^o-\alpha}{2}\;{\color{blue}[1]}

    Since ED = AD,\:\Delta ADE is isosceles: . \angle AED \:=\:\angle A \:=\:\alpha
    . . Then: . \angle ADE \:=\:180^o - 2\alpha\;{\color{blue}[2]}\:\text{ and }\:\angle BED \:=\:180^o-\alpha

    Since BE = ED,\:\Delta BED is isosceles: . \angle EBD = \angle EDB
    . . \angle BED \;+\; \angle EBD \;+\; \angle EDB \:=\:180^o \quad\Rightarrow\quad(180^o-\alpha) \;+\; 2\angle EDB \:=\:180^o . \Rightarrow\quad \angle EDB \:=\:\frac{\alpha}{2}\;{\color{blue}[3]}


    At point D, we have: . \angle ADE + \angle EDB + \angle BDC \:=\:180^o

    Substitute {\color{blue}[2]}, {\color{blue}[3]}, {\color{blue}[1]}: . (180^o - 2\alpha) + \frac{\alpha}{2} + \frac{180^o-\alpha}{2} \:=\:180^o

    Solve for \alpha\!:\;\;\boxed{\alpha \:=\:45^o}


    You are right, jashansinghai!
    .
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