$$ABCD$$ is a parallelogram. The measure of $$\angle ABC = 60^\circ$$. The bisector of $$\angle BAD$$ (dividing $$\angle BAD$$ in two equal halves) meets the side $$DC$$ in point $$E$$ such that $$E$$ is the midpoint of side $$DC$$. If the length of segment $$AE = 6\text{ cm}$$, then find Perimeter of $$\triangle ADE$$ + perimeter of $$\square ABCE$$.

Adjacent angles of a parallelogram are supplementary, so $$\angle BAD = 180^\circ - 60^\circ = 120^\circ$$ and the bisector gives $$\angle BAE = \angle EAD = 60^\circ$$. Since $$AB \parallel DC$$, $$\angle AED = \angle BAE = 60^\circ$$ (alternate angles), so $$\angle ADE = 60^\circ$$ and $$\triangle ADE$$ is equilateral with $$AD = DE = AE = 6\text{ cm}$$. As $$E$$ is the midpoint of $$DC$$, $$EC = 6$$, so $$AB = DC = 12$$ and $$BC = AD = 6$$. Perimeter of $$\triangle ADE = 6 + 6 + 6 = 18\text{ cm}$$ and perimeter of $$ABCE = AB + BC + CE + EA = 12 + 6 + 6 + 6 = 30\text{ cm}$$, so the required sum is $$18 + 30 = 48\text{ cm}$$.