Find the length of the angle bisector

**shiny718** · August 3rd 2012, 05:40 AM

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.

**yamika** · August 5th 2012, 06:45 PM

$Find the length of the angle bisector-math.jpg$

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

**Soroban** · August 6th 2012, 05:52 AM

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     :

$\text{Since the areas of }\Delta ABD\text{ and }\Delta ADC\text{ are in the ratio }2\!:\!3\text{ and they have the same height,}$
. . $\text{their bases are in the ratio }2\!:\!3.\:\text{ That is: }\:BD\!:\!DC \,=\,2\!:\!3$
$\text{Since }BC = 15,\,\text{ then: }\:BD = 6,\;DC = 9.$

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

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

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

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

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

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

. . . $x^2 \:=\:42 \quad\Rightarrow\quad x \:=\:\sqrt{42}$

**richard1234** · August 6th 2012, 08:57 AM

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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