In the figure, AFB and ADB are triangles. ABC, FEB and AED are straight lines. Given that $\angle\text{DEB}$ is $79^\circ$ and $\angle\text{DBC}$ is $85^\circ$, find the value of $\angle\text{DAB} + \angle\text{ADB} + \angle\text{EAF} + \angle\text{AFE}$.

1. Find the sum of $\angle\text{DAB}$ and $\angle\text{ADB}$ using the exterior angle of $\triangle ADB$.

In $\triangle ADB$, the exterior angle at $B$ is $\angle\text{DBC}$.
$\angle\text{DAB} + \angle\text{ADB} = \angle\text{DBC} = 85^\circ$

2. Find $\angle\text{AEF}$ using vertically opposite angles.

Since $AED$ and $FEB$ are straight lines, $\angle\text{AEF}$ and $\angle\text{DEB}$ are vertically opposite angles.
$\angle\text{AEF} = \angle\text{DEB} = 79^\circ$

3. Find the sum of $\angle\text{EAF}$ and $\angle\text{AFE}$ using the angle sum of $\triangle AFE$, and calculate the final total.

In $\triangle AFE$, the sum of angles is $180^\circ$.
$\angle\text{EAF} + \angle\text{AFE} = 180^\circ - \angle\text{AEF} = 180^\circ - 79^\circ = 101^\circ$

Total sum $= (\angle\text{DAB} + \angle\text{ADB}) + (\angle\text{EAF} + \angle\text{AFE})$
$= 85^\circ + 101^\circ = 186^\circ$