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Thread: problem!!!!

  1. #1
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    problem!!!!

    Let ABC be an isosceles triangle with AB = AC. Suppose that the angle bisector
    of angle B meets AC at D and that BC = BD + AD. Determine angle A.
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  2. #2
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    Re: problem!!!!

    Hello, geniusgarvil!

    This reminds me of a classic problem.
    Is there a typo?


    $\displaystyle ABC$ is an isosceles triangle with $\displaystyle AB = AC.$
    The bisector of angle $\displaystyle B$ meets $\displaystyle AC$ at $\displaystyle D$
    and $\displaystyle BC \,=\, BD\;{\color{red}=}\;AD.$
    Determine angle $\displaystyle A.$

    Code:
                  A
                  *
                 / \
                / θ \
               /     \
              /       * D
             /     * 2θ\
            / θ *       \
           / * θ      2θ \
        B * - - - - - - - * C
    Let $\displaystyle \theta = \angle ABD = \angle DBC$

    Since $\displaystyle \Delta ADB$ is isosceles, $\displaystyle \angle A \,=\,\theta$

    $\displaystyle \angle BDC$ is an exterior angle to $\displaystyle \Delta ADB.$
    . . Hence: $\displaystyle \angle BDC \,=\,2\theta$

    Since $\displaystyle \Delta BDC$ is isosceles, $\displaystyle \angle C \,=\,2\theta$

    In $\displaystyle \Delta BDC\!:\;\theta + 2\theta + 2\theta \:=\:180^o \quad\Rightarrow\quad 5\theta \:=\:180^o$

    Therefore: .$\displaystyle \angle A \:=\:\theta \:=\:36^o$
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  3. #3
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    Re: problem!!!!

    Let the base angles be$\displaystyle 2\theta,$ (later, angles in degrees), in which case the angle $\displaystyle BDA=3\theta.$
    Using the sine rule in triangles CBD and ABD, we have, respectively,

    $\displaystyle \frac{BD}{\sin(2\theta)}=\frac{CB}{\sin(3\theta)},$ in which case $\displaystyle BD=\frac{CB\sin(2\theta)}{\sin(3\theta)},$

    and

    $\displaystyle \frac{AD}{\sin\theta}=\frac{BD}{\sin(4\theta)},$ so $\displaystyle AD=\frac{BD\sin\theta}{\sin(4\theta)}=\frac{CB\sin \theta\sin(2\theta)}{\sin(4\theta)\sin(3\theta)}$


    Substitute into the given equation $\displaystyle AD+BD=CB$, and cancel $\displaystyle CB$ throughout,

    $\displaystyle \frac{\sin\theta\sin(2\theta)}{\sin(4\theta)\sin(3 \theta)}+\frac{\sin(2\theta)}{\sin(3\theta)}=1.$

    Multiply throuout by $\displaystyle \sin(4\theta)\sin(3\theta),$

    $\displaystyle \sin\theta\sin(2\theta)+\sin(2\theta)\sin(4\theta)-\sin(4\theta)\sin(3\theta)=0.$

    Now use the trig identity

    $\displaystyle \cos A -\cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

    $\displaystyle \{\cos(3\theta)-\cos\theta\}+\{\cos(6\theta)-\cos(2\theta)\}-\{cos(7\theta)-\cos\theta\}=0.$

    Rearrange,

    $\displaystyle \{\cos(\6\theta)+\cos(3\theta)\}-\{\cos(7\theta)+\cos(2\theta)\}=0,$

    and now use the identity

    $\displaystyle \cos A + \cos B = 2\cos \left( \frac{A+B}{2}\right)\cos \left( \frac{A-B}{2}\right)$

    so

    $\displaystyle 2\cos(9\theta/2)\cos(3\theta/2)-2\cos(9\theta/2)\cos(5\theta/2)=0,$

    $\displaystyle \cos(9\theta/2)\{\cos(3\theta/2)-\cos(5\theta/2)\}=0,$

    and therefore

    $\displaystyle \cos(9\theta/2)=0,$ since $\displaystyle \cos(3\theta/2)-\cos(5\theta/2)\neq 0,$

    Therefore

    $\displaystyle 9\theta/2=90,\quad \theta = 20$ in whch case angle $\displaystyle A=100.$
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  4. #4
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    Re: problem!!!!

    thanks it is really awesome
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