Not For Those Who Are Slaves of Trigonometry

The goal of the solution is to prove that $AD$ is the bisector of $\angle BAC$ (that's why I hide the incenter).

Draw the bisector of $\angle BAC$ , it cuts $BC$ at point $D'$ and cuts $MN$ at point $P'$ . Extend $CP'$ to cut $AB$ at point $K$ . Let us call $I$ is the incenter of triangle $ABC$ . To achieve the goal, first, we'll prove $IEP'C$ is concylic, then, $E, F, P'$ are colinear. If we do so, $P$ and $P'$ both are intersections of $EF$ and $MN$ , so $P \equiv P'$ or $D \equiv D'$ .

It's quite easy to prove $IEP'C$ is concylic. $M$ and $N$ are midpoints of $AC$ and $BC$ , respectively, so $P'$ is midpoint of $CK$ . But $AP'$ is bisector of $\angle BAC$ , triangle $KAC$ is isosceles, and $AP'$ is the altitude of this triangle. $\angle IEC = \angle IP'C = 90^\circ$ , which makes $IEP'C$ concylic.

Now we'll prove the second, $EFP'$ is a straight line. $\angle P'IC$ is the external angle of triangle $AIC$ , so $\angle P'IC = \dfrac 12(\angle BAC + \angle ACB)$ or $\angle P'IC = \dfrac 12(180^\circ - \angle ABC) = \angle BEF$ . But $\angle P'IC = \angle P'EC$ (property of a concylic quadrilateral), we have $\angle P'EC = \angle BEF$ . Yay! $E, F, P'$ is colinear now.

We've just proved that $EF, MN$ and the bisector of $\angle BAC$ are concur. The last part is easy now. $AD$ is now the bisector of $\angle BAC$ so $\frac{DB}{AB}=\frac{DC}{AC}=\frac{DB+DC}{AB+AC}=\dfrac 69$ . This gives us $DB=\dfrac 83, DC=\frac{10}{3}$ . Use this formula to calculate $AD$ :

$AD = \sqrt{AB \times AC - DB \times DC} = \sqrt{4 \times 5 - \dfrac 83 \times \frac{10}{3}} = \frac{10}{3}$ .

Thus, the answer is $\boxed{13}$ .

We can also use the formula for length of angle bisector for angle A That is 2bc.Cos(A/2)/(b+c) putting these values we get AD = 10/3