Let us denote the angles as shown in fig. 26. For triangles $AKT$ and $ATN$ let us apply Law of sines and get:
$$\frac{NT}{TK} = \frac{AN \cdot \sin \alpha_1}{AK \cdot \sin \alpha_2} = \frac{\sin \alpha_1}{\sin \alpha_2}$$
Similarly we can get:
$$\frac{NF}{FP} = \frac{NB \cdot \sin \beta_1}{PB \cdot \sin \beta_2} = \frac{\sin \beta_1}{\sin \beta_2}, \quad \frac{PQ}{QK} = \frac{PC \cdot \sin \gamma_1}{KC \cdot \sin \gamma_2} = \frac{\sin \gamma_1}{\sin \gamma_2}$$
If we multiply all these expressions, we will get:
$$\frac{NT \cdot KQ \cdot PF}{TK \cdot QP \cdot FN} = \frac{\sin \alpha_1 \cdot \sin \beta_2 \cdot \sin \gamma_2}{\sin \alpha_2 \cdot \sin \beta_1 \cdot \sin \gamma_1} = 1$$
as far as $AL$, $BL$ and $CL$ intersect in one single point, and for them Cheva's theorem in its trigonometrical form is fulfilled. But then from Cheva's theorem in its standard form, it follows that $BK$, $NQ$ and $PT$ intersect in one single point, that is what had to be proven.