Let $P$ be the midpoint of segment $BC$ and $J$ be the midpoint of arc $\widearc{BC}$ ($J \neq A$).

We call the circumcircle of triangle $\triangle CID$, $\omega$ and the intersection point of $\omega$ and segment $BC$, $R$. We have
$$\angle QIC = 180^{\circ} - (\angle IQC + \angle ICQ) = 180^{\circ} - (\angle PIC + \angle ICP) = 90^{\circ}.$$
So, $\angle QIC = 90^{\circ}$ and $N$ is the center of $\omega$ which gives us $\angle QRC = 90^{\circ}$ and $QR \parallel AP$. $\angle RDC = \angle RQC$ gives us $\angle JAC = \angle JDC$. So, points $D, R$ and $J$ are collinear. Since $\angle RNC = 2\angle RQC = 2\angle PAC = \angle A$, we have $RN \parallel AB$. Therefore
$$\frac{AN}{NC} = \frac{BR}{RC} \implies \frac{AN}{NC} \cdot \frac{RC}{BR} \cdot \frac{BM}{MA} = 1.$$
So, the lines $AR$, $BN$ and $CM$ are concurrent. Let $X$ be the intersection point of lines $AD$ and $BC$. It suffices to show that $(XR, CB) = -1$. Since $D, R$ and $J$ are collinear and $J$ is the midpoint of arc $\widearc{BC}$, We have
$$(XR, CB) \stackrel{D}{=} (AJ, CB) = -1.$$
Hence the result.