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