A geometry problem by Muhammad Maulana

Apply cosine law on triangle $ABC$ .

$(BC)^2=(AB)^2+(AC)^2-2(AB)(AC)(\cos A)$

$8^2=7^2+9^2-2(7)(9)(\cos A)$

$A \approx 58.41$

Since triangle $ABD$ is isosceles, $\angle A=\angle BDA$ , so $\angle ABD=180 - 2\times 58.41=63.18$

By sine law on triangle $ABD$ ,

$\dfrac{AD}{\sin 63.18}=\dfrac{7}{\sin 58.41}$

$AD \approx \boxed{7.33}$

Applying cosine law to Tr.ABC ---> cosB = 2/7 ---> sinB = 3sqrt5 /7

Applying sine law ---> 9/sinB = 8/sinA = 7/sinC

sinA = 8sqrt5 /21 , sinC = sqrt5 /3 ---> cosA = 11/21 , cosC = 2/3

m<DBC = m<ADB - m<C = m<A - m<C

cos(DBC) = cos(A-C) = 62/63

DC^2 = 7^2 + 8^2 - (2) (7) (8)*(62/63) = 25/9

DC = 5/3 ---> AD = 9 - 5/3 = 22/3 = 7.333333