Proof Through point $L$ draw perpendicular lines to $AB$ and to $AC$, the feet are $L'$ and $L''$ respectively. Let $\angle LAB = \alpha_1$, $\angle LAC = \alpha_2$, $\angle LF_2A = \alpha_3$, and $\angle LE_1A = \alpha_4$. We have
$$\frac{\sin \alpha_1}{\sin \alpha_2} = \frac{LL'}{LL''} = \frac{LF_2 \sin \alpha_3}{LE_1 \sin \alpha_4}. \quad ①$$
Draw line segments $D_1F_2$ and $D_2E_1$ (see Figure 1). Since $\triangle LD_1F_2 \sim \triangle LD_2E_1$ we get

Figure 1

Figure 2
$$\frac{LF_2}{LE_1} = \frac{D_1 F_2}{D_2 E_1}. \qquad \textcircled{2}$$
Draw line segments $D_2F_1$ and $D_1E_2$ (see Figure 2). By using the sine rule we obtain
$$\frac{\sin \alpha_3}{\sin \alpha_4} = \frac{D_2 F_1}{D_1 E_2}. \qquad \textcircled{3}$$
Substituting ② and ③ into ①, we have
$$\frac{\sin \alpha_1}{\sin \alpha_2} = \frac{D_1 F_2}{D_2 E_1} \cdot \frac{D_2 F_1}{D_1 E_2}. \qquad \textcircled{4}$$
Similarly, write $\angle BMC = \beta_1$, $\angle MBA = \beta_2$, $\angle NCA = \gamma_1$, $\angle NCB = \gamma_2$, and we get
$$\frac{\sin \beta_1}{\sin \beta_2} = \frac{E_1 D_2}{E_2 F_1} \cdot \frac{E_2 D_1}{E_1 F_2}, \qquad \textcircled{5}$$
$$\frac{\sin \gamma_1}{\sin \gamma_2} = \frac{F_1 E_2}{F_2 D_1} \cdot \frac{F_2 E_1}{F_1 D_2}. \qquad \textcircled{6}$$
Multiplying ④, ⑤ and ⑥, we obtain
$$\frac{\sin \alpha_1}{\sin \alpha_2} \cdot \frac{\sin \beta_1}{\sin \beta_2} \cdot \frac{\sin \gamma_1}{\sin \gamma_2} = 1.$$
Finally, according to the inverse of Ceva's Theorem, we know $AL$, $BM$ and $CN$ have a common point.