AEF and CEF are isosceles triangles with AE = EF = CE. Find $\angle\text{AFB}$.

1. Find $\angle\text{EFC}$ using the isosceles triangle CEF.

In $\triangle CEF$, since $CE = EF$, it is an isosceles triangle.
$\angle\text{EFC} = \angle\text{FCE} = 27^\circ$.
Since $F, B, C$ are collinear, $\angle\text{EFB} = 27^\circ$.

2. Find $\angle\text{FEB}$ using the exterior angle of $\triangle EFB$.

In $\triangle EFB$, the exterior angle at $B$ is $\angle\text{ABF}$.
$\angle\text{FEB} + \angle\text{EFB} = \angle\text{ABF}$
$\angle\text{FEB} + 27^\circ = 85^\circ$
$\angle\text{FEB} = 85^\circ - 27^\circ = 58^\circ$.
Since $A, B, E$ are collinear, $\angle\text{AEF} = 58^\circ$.

3. Find $\angle\text{AFE}$ using the isosceles triangle AEF.

In $\triangle AEF$, since $AE = EF$, it is an isosceles triangle.
$\angle\text{AFE} = \frac{180^\circ - \angle\text{AEF}}{2}$
$\angle\text{AFE} = \frac{180^\circ - 58^\circ}{2} = \frac{122^\circ}{2} = 61^\circ$.

4. Calculate $\angle\text{AFB}$.

Since $B$ lies on $AE$ and $FC$, the ray $FB$ is between $FA$ and $FE$.
$\angle\text{AFB} = \angle\text{AFE} - \angle\text{EFB}$
$\angle\text{AFB} = 61^\circ - 27^\circ = 34^\circ$.