Using the condition of the problem, we get
$$\angle BAE = \angle BAC + \angle CAE = \angle AEC + \angle CAE$$
(Fig. 22). From the triangle $ACE$ we get $\angle AEC + \angle CAE = \angle DCE$. Thus $\angle BAE = \angle DCE$. At the same time, $AE = CD$ and $AB = CE$. Consequently, the triangles $AEB$ and $CDE$ are equal, hence the perimeter of the triangle $AEB$ is $35$ cm.

Fig. 22

Now we get
$$\begin{aligned} EA+AB+BC+CD+DE &= (EA+AB+BE-CE) + (CD+DE+EC-CE) \\ &= (EA+AB+BE) + (CD+DE+EC) - 2CE. \end{aligned}$$
Since $EA + AB + BE = CD + DE + EC = 35 \text{ cm}$ and $CE = 8 \text{ cm}$, the perimeter of the pentagon $ABCDE$ is $2 \cdot 35 \text{ cm} - 2 \cdot 8 \text{ cm} = 54 \text{ cm}$.