Angle bisector problem

There is a theorem that AB $\times$ AC = $AD^{2}$ + BD $\times$ DC

If you want a proof then comment below and I will add one.

Now, applying the theorem, we get

8 $\times$ 6 = $AD^{2}$ + 4 $\times$ 3

48 = $AD^{2}$ + 12

$AD^{2}$ = 36

AD = 6cm

Hence the answer is 6cm

We can do this by direct formula ,

$t_{BC}=\frac{2\sqrt{\overline{AB}\cdot \overline{AC}\cdot s(s-\overline{BC})}}{\overline{AB}+\overline{AC}}$

Here ,

$t_{BC}\rightarrow$ Angle bisector on $\overline{BC}$ .

$s\rightarrow$ Semi-perimeter.

After calculation, Answer = $\boxed{6}$

For the formula, go here: https://proofwiki.org/wiki/Length of Angle_Bisector

Stewarts theorem gives AD^2=(8 8 3+6 6 4)/(4+3)-4*3=36. AD=6.