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Thread: 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 $\displaystyle ABC$ where $\displaystyle AB=AC$.
    $\displaystyle D$ is on $\displaystyle AC$ such that $\displaystyle BC=BD$.
    $\displaystyle E$ is on $\displaystyle AB$ such that $\displaystyle BE=ED=AD$.

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

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

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

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

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


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

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

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


    You are right, jashansinghai!
    .
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