jmc

Since $AE$ and $AF$ are tangents from the same point to the same circle, $AE=AF$. Let $x=AE=AF$. Similarly, let $y=BD=BF$ and $z=CD=CE$.



Then $x+y=AB=13$, $x+z=AC=15$, and $y+z=BC=14$. Adding all these equations, we get $2x+2y+2z=42$, so $x+y+z=21$. Subtracting the equation $y+z=14$, we get $x=7$.

By Heron's formula, the area of triangle $ABC$ is $$K=\sqrt{21(21-14)(21-15)(21-13)}=84,$$so the inradius is $r=K/s=84/21=4$.

We can divide quadrilateral $AEIF$ into the two right triangles $AEI$ and $AFI$.



The area of triangle $AEI$ is $$\frac{1}{2}\cdot AE\cdot IE=\frac{1}{2}\cdot x\cdot r=\frac{1}{2}\cdot 7\cdot 4=14,$$and the area of triangle $AFI$ is also 14. Therefore, the area of quadrilateral $AEIF$ is $14+14=\boxed{28}$.