Show that $PM=MS$

Question

In a triangle $ABC$ point $M$ is the middle of $AB$. On the line $CM$ we choose an arbitrary point $N$, that lies outside the triangle on the other side of $AB$ than $C$. We draw a line through $N$ that intersects the lines $AB$ and $AC$ in points $P$ and $Q$. Let $R$ be the intersection of lines $QM$ and $BN$. Let $S$ be the intersection of $AB$ and $CR$. Show that $MP=MS$.

Answer 1

Note that this problem is completely projective, and thus any quadrilateral can be projected to any other quadrilateral and cross-ratios remain preserved. Let $P_{\infty}$ be the point at infinity along $AB$. Thus, $(A, B; M, P_{\infty}) = -1$. We take the projection that sends $CBNA$ to square $C'B'N'A'$, and label the other points in question in a similar manner.

Since $M = CN \cap AB$, $M' = C'N' \cap A'B'$, i.e. the midpoint of $AB$. Thus, $P'_\infty$ is still at infinity.

We will show $(P', S'; M', P'_{\infty}) = -1$. Since $Q'R'$ passes through $M'$, $M'$ must bisect $Q'R'$ by symmetry of the square. We also have $C'M' = M'N'$ and $\angle C'M'R' = \angle N'M'Q'$. Thus, $\triangle C'M'R' \cong \triangle N'M'Q'$.

Finally, since $\angle Q'M'P' = \angle R'M'S'$, $P'M'=M'S'$, so $(P', S'; M', P'_{\infty}) = -1$. Thus, proven.