Let $NY$ and $NZ$ intersect $BC$ at $E$ and $F$, respectively. Note that since quadrilaterals $ABYN$ and $ACZN$ are cyclic,
$$\angle NZY + \angle NZY = \angle NAB + \angle NAC = \angle A \implies \angle ZNY = 180^{\circ} - \angle A,$$
hence the pentagon $ACZNE$ is cyclic. Similarly one can show that $ABYNF$ is cyclic, too. Now it is easy to conclude that $\triangle BEY \sim \triangle CFZ$.

Let $P$ be the intersection of $\ell_b$ and $\ell_c$. Now by law of sines
$$\frac{\sin \angle PNZ}{\sin \angle PNY} = \frac{PZ \cdot \sin \angle NZP}{PY \cdot \sin \angle NYP}.$$

So we have to prove
$$\frac{PY}{PZ} = \frac{\sin \angle NZP}{\sin \angle NYP} \iff \frac{PY}{PZ} = \frac{\sin \angle CZF}{\sin \angle BEY}.$$
In order to prove this, note that by Menelaus's theorem one can have
$$\frac{PY}{YB} \cdot \frac{BX}{XC} \cdot \frac{CZ}{ZP} = 1 \implies \frac{PY}{PZ} = \frac{CX}{BX} \cdot \frac{BY}{ZC} = \frac{AC}{AB} \cdot \frac{BY}{CZ}.$$
Hence we need to have
$$\begin{align*} \frac{AC}{AB} \cdot \frac{BY}{CZ} &= \frac{\sin \angle CZF}{\sin \angle BYE} &\iff& \frac{AC}{AB} \cdot \frac{BY}{\sin \angle CZF} \cdot \frac{\sin \angle BYE}{CZ} &= 1 \\ &\iff& \frac{AC}{AB} \cdot \frac{BE}{CZ} &= 1 \\ &\iff& \frac{AC}{AB} \cdot \frac{AB}{AC} &= 1, \end{align*}$$
as desired.