Let $AD$ meet $BE$ at $X$, $BE$ meet $CF$ at $Y$, and $CF$ meet $AD$ at $Z$. Since $AB \parallel ED$, we know that $\triangle XAB \sim \triangle XDE$. Also, the midpoints $H$ and $K$ of $AB$ and $DE$ are corresponding points under this similarity. Thus, $X$ lies on $HK$. Similarly, $Y$ lies on $IL$, and $Z$ lies on $JM$.

By Ceva's theorem, it suffices to prove
$$\frac{\sin \angle AXH}{\sin \angle HXB} \times \frac{\sin \angle EYL}{\sin \angle LYF} \times \frac{\sin \angle CZJ}{\sin \angle JZD} = 1. \qquad (1)$$
Firstly, since $AH = HB$ and $\triangle XAB \sim \triangle XDE$, we have
$$\frac{\sin \angle AXH}{\sin \angle HXB} = \frac{XB}{XA} = \frac{BE}{AD}.$$
By symmetry, we have $\frac{\sin \angle EYL}{\sin \angle LYF} = \frac{CF}{BE}$ and $\frac{\sin \angle CZJ}{\sin \angle JZD} = \frac{AD}{CF}$. Therefore, the left-hand side of (1) is
$$\frac{BE}{AD} \times \frac{CF}{BE} \times \frac{AD}{CF} = 1$$
as desired.