The base $A D$ of a trapezoid $A B C D$ measures $30 \mathrm{~cm}$. Suppose there is a point $E$ on the base $A D$ such that the triangles $\triangle A B E, \triangle B C E$ and $\triangle C D E$ have equal perimeters. Determine the length of $B C$.
We want to prove that $A E$ is equal to $B C$. For this, suppose that $A E$ is greater than $B C$ and choose the point $A^{\prime}$ on $A E$ such that $E A^{\prime}=B C$. By construction, $E A^{\prime}$ and $B C$ are parallel, so $A^{\prime} B C E$ is a parallelogram and, in particular,

$$
A^{\prime} B=C E
$$

By the triangle inequality, we have

$$
A^{\prime} A+A B>A^{\prime} B
$$

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-109.jpg?height=277&width=665&top_left_y=2100&top_left_x=1164)

$$
\begin{aligned}
E A+A B+B E & =E A^{\prime}+A^{\prime} A+A B+B E \\
& >E A^{\prime}+A^{\prime} B+B E=B C+C E+E B
\end{aligned}
$$

From this, it follows that the perimeter of triangle $\triangle A B E$ is greater than the perimeter of triangle $\triangle B C E$, contrary to the given problem.

Through this contradiction, we establish that, given the hypotheses of the problem, $A E$ cannot be greater than $B C$. By a completely analogous process, we can also establish that, conversely, $B C$ cannot be greater than $A E$, from which we conclude that $B C=A E$. The same reasoning can be used to show that $B C=E D$. Thus,

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