Eccentric givens!

By stewart's theorem,

$(BC)((AD)^2+(BD)(DC)) = (AB)^2(DC)+(AC)^2(BD)$

By the angle bisector theorem,

$\frac{AB}{BD} = \frac{AC}{CD}$ or $(AB)(DC) = (AC)(BD)$

Substituting this into the above equation gives us

$(BC)((AD)^2+(BD)(DC)) = (AB)(AB)(DC)+(AC)(AC)(BD) = (AB)(AC)(BD)+(AB)(AC)(DC) = (AB)(AC)((BD)+(CD)) = (AB)(AC)(BC)$ (Since $BD+CD = BC$ )

Dividing both sides by $BC$ gives us

$(AD)^2+(BD)(DC) = (AB)(AC)$

Substituting the given values of $(BD)(DC)$ and $(AB)(AC)$ gives us

$(AD)^2+6 = 10$ , $(AD)^2 = 4$ , $AD = \boxed{2}$

BY ANGLE BISECTOR THEOREM ,

(AD)^2 = AB●AC - BD●CD

→ (AD)^2 = 10 - 6 = 4

→ (AD) = 2 .

its quite simple. !

U can refer to Angle bisector theorem wiki . Thank you.

This is another problem which can be tackled using sine rule(Atleast I like to approach side-length problems using this principle)

For any $\bigtriangleup$ $ABC$ , we have $\displaystyle\frac{\sin\angle\,A}{a}=\frac{\sin\angle\,B}{b}=\frac{\sin\angle\,C}{c}$

Let $\angle$ $BAD=\angle$ $DAC=\alpha,\angle$ $ADB=\beta$ . Then using sine-rule in $\bigtriangleup$ $ABD$ & $ADC$ we have

$\displaystyle\frac{\sin\alpha}{BD}=\frac{\sin\beta}{AB}=\frac{\sin\left(\alpha+\beta\right)}{AD}\;\;\;\;-\left(i\right)$

$\displaystyle\frac{\sin\alpha}{DC}=\frac{\sin\left(\pi-\beta\right)}{AC}=\frac{\sin\left(\beta-\alpha\right)}{AD}\;\;\;-\left(ii\right)$

Now multiplying $\left(i\right)$ & $\left(ii\right)$ respectively, we get

$\displaystyle\frac{\sin^2\alpha}{BD\cdot\,DC}=\frac{\sin^2\beta}{AB\cdot\,AC}=\frac{\sin^2\beta\cos^2\alpha-\sin^2\alpha\cos^2\beta}{AD^2}$

$\implies\displaystyle\frac{\sin^2\alpha}{6}=\frac{\sin^2\beta}{10}=\frac{\sin^2\beta\cos^2\alpha-\sin^2\alpha\cos^2\beta}{AD^2}\;\;\;-\left(*\right)$

From the first relation we have $\displaystyle\frac{\sin^2\alpha}{\sin^2\beta}=\frac{3}{5}\;\;\;-\left(iii\right)$

From last two terms of $\left(*\right)$ & using $\left(iii\right)$ we get, after simplifying $AD=\boxed{2}$ .

$\text{By the angle bisector theorem,}\\ BD=\dfrac{BC}{AB+AC}*AB,\ \ and\ \ DC=\dfrac{BC}{AB+AC}*AC,\\ \therefore\ BD*DC=\left (\dfrac{BC}{AB+AC} \right)^2*(AB*AC)....\color{#3D99F6}{***}\\ \text{But by stewart's theorem,}\\ AD^2=(AB*AC)*\left (1- \left (\dfrac{BC}{AB+AC} \right )^2 \right )=(AB*AC) - \left (\dfrac{BC}{AB+AC} \right)^2*(AB*AC)\color{#3D99F6}{***}\\ \therefore\ \ AD=\sqrt{10-6}=2.$ .