Let $N$ be the midpoint of segment $S_1S_2$. First see that sines law on $\triangle APC$ tells us that $\frac{AC}{sin \angle APC}=2r_1$, similarly on $\triangle BPC$ $\frac{BC}{sin \angle BPC}=2r_2$, but $sin \angle APC= sin \angle BPC$ then we get $\frac{AC}{2r_1}=\frac{BC}{2r_2}$, from here that $\triangle AS_1C \sim \triangle BS_2C$. Now there is an obvious spiral similarity that takes $\triangle AS_1C$ to $\triangle BS_2C$, furthermore note that $M, N$ are midpoints of the lines that join the corresponding vertices of these triangles, so we actually have that $\triangle AS_1C \sim \triangle MNC \sim \triangle BS_2C \implies NM=NC$ and the result follows.

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