Law of tangents with an angle bisector without knowing angles "In $\triangle ABC$, let D be a point on BC such that AD bisects $\angle A$. If AD=6, BD=4, and DC=3, then find AB"
This problem is from the Mu Alpha Theta 1991 contest, it appears in volume 2 of AoPS and involves law of tangents. I have tried looking the possible cases where I can use law of tangents, but I don´t see clear the path to solve the problem. So I hope you can give me a hint to be able to solve this problem. Thanks in advance for your help.
 A: 
Given $a_1=|BD|=4$, $a_2=|DC|=3$
and $d=|AD|=6$, let $|AC|=b$ and $|AB|=c$.
By the Angle bisector theorem
for $\angle CAB$ we can express $c$ in terms of $b$:
\begin{align}
c &= \frac{a_1}{a_2}\cdot b
,
\end{align}
and by the
Stewart’s Theorem
we have one more equation in terms of $b$ and $c$:
\begin{align}
a_1 b^2+a_2 c^2&=(a_1+a_2)(d^2+a_1a_2)
,
\end{align}
so we can find $b^2$ as
\begin{align}
b^2 &= \frac{a_2}{a_1}\cdot(d^2+a_1a_2)
=36
,
\end{align}
hence,
\begin{align}
b&=6
,\quad
\text{and }\quad
c=8
.
\end{align}
A: Geometric solution:

Conidering figure we have:
$AK=AD+DK=6-DK$
$AH=AD-DH=6-DH$
In triangle CC'E: $CD=DE\Rightarrow BE=4-3=1$
Triangles BC'E and BAD are similar and we have:
$\frac{BE}{BD}=\frac {C'E}{AD}\Rightarrow C'D=\frac 64=\frac 32$
$\frac{DH}{C'D}=\frac 36=\frac 12\Rightarrow DH=\frac12 C'D=\frac34$
Triangles DHC and DKB are similar so:
$DH=\frac43 DH \Rightarrow DK=1$
In right angle triangle ABK we have:
$AK=AD+DK=6+1=7$
$BK^2=4^2-1^2=15$
$AB=\sqrt{BK^2+AK^2=15+7^2=64}=8$
A: Let's denote AB=c , BC=a and AC=b
Therefore , BC=BD+CD=7
a=7
Now , as AD is the angle bisector ,
Therefore , AB/AC = BD/CD ,or
c/b = 4/3 = k(say)
Therefore , c=4k , b=3k
Now , the length of angle bisector AD is given by
AD=(2bc/(b+c))cos(A/2)
Plugging in AD=6 , b=3k and c=4k , we get the value of cos(A/2) as
cos(A/2)=7/4k
Now , calculating value of cos(A)
cos(A)= 2cos²(A/2)-1
Therefore ,
cos(A)=(49-8k²)/8k²
Now , using cosine rule ,which states that ,
cos(A)= (b²+c²-a²)/2bc
Plugging in values
49-8k²/8k² = {(4k)² + (3k)²- a²}/2(4k)(3k)
=>49(3)-24k²=25k²-49
49(4)=49k², therefore , k²=4
Therefore , k=2 (neglecting negative value )
Therefore , AB=4k=4(2)=8
