Let $MP$ and $NQ$ intersect $AD$ at $K_1$ and $K_2$ respectively. By applying Menelaus' theorem to triangle $ABD$ and line $MPK_1$, we have
$$\frac{AK_1}{K_1D} = \frac{AP}{PB} \cdot \frac{BM}{MD} = \frac{AP}{2PB}$$
and similarly $\frac{AK_2}{K_2D} = \frac{AQ}{2QC}$. A homothety at $A$ takes $\Gamma \to \omega$ and $D$ to the midpoint of arc $BC$ not containing $A$, so $PQ \parallel BC$ and $AD$ bisects $BAC$. Thus
$$\frac{AK_1}{K_1D} = \frac{AP}{2PB} = \frac{AQ}{2QC} = \frac{AK_2}{K_2D}$$
which implies $K_1 = K_2$, and $K$ lies on $AD$. Then we obtain
$$\angle JXD = \angle JXK = \angle JIK = \angle JIP = \angle JQP = \angle JND$$
where the last equality follows from $PQ \parallel BC$. This shows $JXND$ is cyclic and hence
$$\angle DXN = \angle DJN = \angle DJQ = \angle DAQ = \angle DAC$$
which shows $AC \parallel XN$. As $C$ is the midpoint of $DN$, $A$ is the midpoint of $XD$.

Now observe that $\angle ADP = \angle AQP = \angle ACB$ and $\angle PAD = \angle DAC = \frac{\angle A}{2}$ so triangle $APD$ and $ADC$ are similar. Therefore we have
$$\frac{CD}{DP} = \frac{AD}{AP} = \frac{XA}{AP}$$
and also have
$$\angle CDP = 180^\circ - \angle PDB = 180^\circ - \angle PAD = \angle XAP.$$
Combining the two results gives triangles $PDC$ and $PAX$ are similar, which shows $P$ is the centre of spiral similarity taking $CD \to XA$. Hence also triangles $PXC$ and $PAD$ are similar which shows $\angle PXC = \angle PAD = \frac{\angle A}{2}$. This gives
$$\angle BXP = \angle BXC - \angle PXC = \angle BXC - \frac{\angle A}{2}$$
which is symmetric in $B, C$ giving the result.