Estonian Mathematical Olympiad

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