37th Iranian Mathematical Olympiad

Let $P$ be the midpoint of segment $BC$ and $J$ be the midpoint of arc $\text{widearc}BC$ ($J\ne A$). We call the circumcircle of 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 $$\frac{AN}{NC}=\frac{BR}{RC}⟹\frac{AN}{NC}\cdot \frac{RC}{BR}\cdot \frac{BM}{MA}=1.$$ 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 $\text{widearc}BC$, We have $$(XR,CB)\stackrel{D}{=}(AJ,CB)=-1.$$ Hence the result.