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Math Help - Find the length of the angle bisector

  1. #1
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    Find the length of the angle bisector

    ABC is a triangle with AB=8cm and BC=15cm. The line AD bisects the angle \angle BAC
    and intersects BC at D. Given that the area of \triangle ABD : area of \triangle ADC =2:3,
    find the length of AD.
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  2. #2
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    Re: Find the length of the angle bisector

    Find the length of the angle bisector-math.jpg

    from area ratio, tri ABD/tri ACD = 2/3
    BD/(15-BD) =2/3
    BY solving this BD = 6

    From the Sine Rule for ABD triangle 6/SinA = AD/SinB
    SinB/SinA = AD/6------------Eqn 1

    From the Sine Rule for ABD triangle 6/Sin(A)=8/Sin(A+180-(2A+B))
    3/SIn(A)= 4/Sin(180-(A+B))
    3/Sin(A)= 4/Sin(A+B)
    4Sin(A) = 3Sin(A+B)
    4Sin(A) =3[Sin(A)Cos(B) + Cos(A)Sin(B)]
    Dividing by SinA , 4=3(CosB +CosASinB/SinA)
    Applying from eqn 1; 4 = 3Cos(B) +3Cos(A)*AD/6
    8= 6Cos(B) + AD*Cos(A) -----------Eqn 2

    From the Sine Rule for triangle ACD 9/Sin(A) = AD/Sin(180-(2A+B))
    9/Sin(A) = AD/Sin(2A+B)
    9*Sin(2A+B) = AD* Sin(A)
    9*[Sin(2A)Cos(B) +Cos(2A)Sin(B)] = AD * Sin(A)
    because Sin(2A) = 2Sin(A)Cos(A) and Cos(2A) = Cos^2(A)-Sin^2(A)
    9*[2SIn(A)Cos(A)Cos(B) + Sin(B)*(Cos^2(A)-Sin^2(A))] = AD*Sin(A)
    18Sin(A)Cos(A)Cos(B) + 9Sin(B)*(Cos^2(A)-Sin^2(A)) = AD*Sin(A)
    Dividing both sides by Sin(A);
    18Cos(A)Cos(B) + 9(Cos^2(A)-Sin^2(A))*Sin(B)/Sin(A) = AD
    Applying from eqn 1 18Cos(A)Cos(B) + 9*AD*(Cos^2(A)-Sin^2(A))/6 = 2AD
    36Cos(A)Cos(B) + 3*AD*Cos^2(A) -3*AD*Sin^2(A) =2AD
    36Cos(A)Cos(B) + 3*AD*[Cos^2(A)-(1-Cos^2(A))] =2AD
    36Cos(A)Cos(B) + 3*AD*[2Cos^2(A)-1] = 2AD
    36Cos(A)Cos(B) + 6*AD*Cos^2(A) = 5AD
    6Cos(A)[6Cos(B)+AD*Cos(A)] = 5AD
    From eqn 2
    6Cos(A)*8 =5AD
    48Cos(A) =5*AD ----------------eqn 3

    Applying the cosine Rule for triangle ABD;
    6^2 = 8^2 +AD^2 -2*AD*8*Cos(A)
    36 = 64 + AD^2 -16AD* Cos(A)
    16AD Cos(A) = AD^2 + 28
    From eqn 3
    16*AD*5*AD/48 = AD^2 +28
    5*AD^2 = 3*AD^2 + 84
    2*AD^2 =84
    AD = Sqrt 42
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  3. #3
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    Re: Find the length of the angle bisector

    Hello, shiny718!

    ABC\text{ is a triangle with }AB=8\text{ cm and }BC=15\text{ cm.}

    \text{The line }AD\text{ bisects }\angle BAC\text{ and intersects }BC\text{ at }D.

    \text{Given that (area of }\Delta ABD) : \text{(area of }\Delta ADC) \;=\;2:3,\,\text{find the length of }AD.

    Code:
                                  A
                                  *
                              *  * *
                   12     *     *   *
                      *        *x    * 8
                  *           *       *
              *            θ * π-θ     *
        C * - - - - - - - - * - - - - - * B
          :        9        D     6     :
    \text{Since the areas of }\Delta ABD\text{ and }\Delta ADC\text{ are in the ratio }2\!:\!3\text{ and they have the same height,}
    . . \text{their bases are in the ratio }2\!:\!3.\:\text{ That is: }\:BD\!:\!DC \,=\,2\!:\!3
    \text{Since }BC = 15,\,\text{ then: }\:BD = 6,\;DC = 9.

    \text{Theorem: an angle bisector divides the opposite side in proportion to the other two sides.}
    . . \text{Hence: }\:\tfrac{AC}{8} \,=\,\tfrac{9}{6}\quad\Rightarrow\quad AC = 12

    \text{Let }x\text{ = length of the angle bisector, }AD.
    \text{Let }\theta \,=\,\angle ADC \quad\Rightarrow\quad \pi-\theta \,=\,\angle ADB


    \text{In }\Delta ADC,\text{ Law of Cosines: }\:\cos\theta \:=\:\frac{x^2+9^2-12^2}{2(9)(x)} \:=\:\frac{x^2-63}{18x}\;\;{\color{blue}[1]}


    \text{In }\Delta ADB,\text{ Law of Cosines: }\:\cos(\pi-\theta) \:=\:\frac{x^2 + 6^2 - 8^2}{2(6)(x)} \:=\:\frac{x^2 - 28}{12x}
    . . \text{This becomes: }\:-\cos\theta \:=\:\frac{x^2 - 28}{12x} \quad\Rightarrow\quad \cos\theta \:=\:\frac{28-x^2}{12x}\;\;{\color{blue}[2]}


    \text{Equate }{\color{blue}[1]}\text{ and }{\color{blue}[2]}:\;\frac{x^2-63}{18x} \:=\:\frac{28-x^2}{12x}

    \text{Multiply by }36x\!:\;2x^2 - 126 \:=\:84 - 3x^2 \quad\Rightarrow\quad 5x^2 \:=\:210

    . . . x^2 \:=\:42 \quad\Rightarrow\quad x \:=\:\sqrt{42}
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  4. #4
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    Re: Find the length of the angle bisector

    A simpler solution: We know that BD = 6 and DC = 9. By an angle bisector theorem,

    \frac{AC}{9} = \frac{8}{6} \Rightarrow AC = 12.

    Now we apply Stewart's theorem on triangle ABC:

    8^2(9) + 12^2(6) = 15(AD^2 + 6*9)

    Solving, we get AD = \sqrt{42}.
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