4. In pentagon $A B C D E A B=B C=C D=D E, \angle B=96^\circ$ and $\angle C=\angle D=108^\circ$. Find angle $E$.
Answer: $102^{\circ}$. Solution. Draw segments $B D$ and $C E$. Let them intersect at point $O$. Note that triangles $B C D$ and $C D E$ are isosceles with an angle of $108^{\circ}$ at the vertex, so the base angles are $36^{\circ}$ (they are marked on the diagram with one arc). Then $\angle B C E = \angle B D E = 72^{\circ}$. Angle $C O D$ is $108^{\circ}$ (since in triangle $C O D$ there are two angles of $36^{\circ}$). Therefore, $\angle C O B = 180^{\circ} - 108^{\circ} = 72^{\circ}$. Angles of $72^{\circ}$ are marked on the diagram with two arcs. We get that triangles $C B O$ and $D E O$ are isosceles. Thus, $A B = B O = B C = C D = D E = E O = x$. Note that $\angle O B A = 96^{\circ} - 36^{\circ} = 60^{\circ}$. Therefore, triangle $O B A$ is isosceles with an angle of $60^{\circ}$ at the vertex, i.e., equilateral. Hence, $A O = x$. Calculate angle $A O E$ $\angle A O E = \angle E O B - \angle A O B = 108^{\circ} - 60^{\circ} = 48^{\circ}$. Triangle $A O E$ is isosceles with an angle of $48^{\circ}$ at the vertex. Therefore, $\angle O E A = (180^{\circ} - 48^{\circ}) / 2 = 66^{\circ}$. We get that angle $E$ of the pentagon is $\angle A E D = \angle A E O + \angle O E D = 66^{\circ} + 36^{\circ} = 102$.

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