The base $A D$ of a trapezoid $A B C D$ measures $30 \mathrm{~cm}$. Suppose there is a point $E$ on $A D$ such that the triangles $\triangle A B E, \triangle B C E$ and $\triangle C D E$ have equal perimeters. Determine the length of $B C$.
Suppose that $A E$ is greater than $B C$, and let $A^{\prime}$ be a point on $A E$ such that $A^{\prime} E = B C$.

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Since $A^{\prime} E$ and $B C$ are parallel, $A^{\prime} B C E$ is a parallelogram, in particular $B A^{\prime} = C E$. But, $A A^{\prime} + A B > B A^{\prime}$ by the triangle inequality. Thus:

$A B + A E + B E = A B + A A^{\prime} + A^{\prime} E + B E > B A^{\prime} + A^{\prime} E + B E = B C + C E + E B$.

Therefore, the perimeter of triangle $\triangle A B E$ is greater than the perimeter of triangle $\triangle B C E$. Hence, $A E$ cannot be greater than $B C$.

By a similar process, we can also conclude that $B C$ cannot be greater than $A E$, and therefore $B C = A E$.

Similarly, we have $E D = B C$. Consequently,

$$
B C = \frac{1}{2}(A E + E D) = 15 \text{ cm}
$$