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