Much easier than P2 OOPS From the triangle inequality, $AN+BM>AB$, so points $A$, $M$, $N$, and $B$ lie on $AB$ in that order. This implies $S$ and $C$ lie on opposite sides of $AB$. Now observe \[ \angle CNA = \angle ACN = 180^\circ - \angle CNS, \]so $NC$ is the external angle bisector of $\angle CNM$. Analogously, $MC$ is also an external angle bisector, so $C$ is the $S$-excenter of $\triangle SMN$. The result follows.

kfas wrote: On side $AB$ of a scalene triangle $ABC$ there are points $M$, $N$ such that $AN=AC$ and $BM=BC$. The line parallel to $BC$ through $M$ and the line parallel to $AC$ through $N$ intersect at $S$. Prove that $\measuredangle{CSM} = \measuredangle{CSN}$. So this is rather clear: As if we take $\angle ACN=\angle CNA=\alpha$ and symmetrically $\angle BCM=\angle CMB=\beta$, it is apparent that $\angle SMN=\pi-2\beta$, and can see $C$ is clearly excenter of $SMN$ due to similarity on the $\alpha$ side. The result is thus too clear.

Let $CS\cap AB=T$ Consider that $\dfrac{NT}{TM}=\dfrac{AN}{BM}=\dfrac{AC}{BC}=\dfrac{SN}{MS}$ Which means that $\angle MSC=\angle CSN$