Solution:

Take $E_1$ on the line $AD$ so that $AE_1CB$ is a parallelogram. Then $AE_1 = BC$, $AB = CE_1$, so triangles $ABE_1$ and $BCE_1$ have equal perimeters. Moreover, $E_1$ is the only point on the line for which this is true. For if we move $E$ a distance $x$ from $E_1$, then we change $AE_1$ by $x$, and $CE_1$ by less than $x$. $AB$ and $BC$ are unchanged. So $AB + BE_1$ and $BC + CE_1$ are changed by different amounts. Hence the perimeters of $ABE_1$ and $BCE_1$ are no longer equal.

Similarly, let $E_2$ be the point on the line $AD$ so that $BCDE_2$ is a parallelogram. Then $E_2$ is the unique point such that $BCE_2$ and $CDE_2$ have equal perimeters. So if all three triangles have equal perimeters, then $E_1$ and $E_2$ must coincide and hence $BC = AE = DE$, so $BC = AD / 2$.