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Let $T = EH \cap IF$; then it suffices to show that $C, T, S$ are collinear, which is equivalent to $\frac{\sin \angle{HTS}}{\sin{\angle{ITS}}} = \frac{\sin \angle{ETC}}{\sin \angle{FTC}}$. Note that by the Law of Sines in $\triangle{THS}, \triangle{TIS}$, $$\frac{\sin \angle{HTS}}{\sin \angle{ITS}} = \frac{\sin \angle{THS} \cdot \frac{HS}{TS}}{\sin \angle{TIS} \cdot \frac{IS}{TS}} = \frac{\sin \angle{THS}}{\sin \angle{TIS}} \cdot \frac{\sin \angle{FAC}}{\sin \angle{EBC}} = \frac{\sin \angle{THS}}{\sin \angle{TIS}} \cdot \frac{EC}{FC} \cdot \frac{AF}{BE}.$$Furthermore, by the Law of Sines in $\triangle{TEC}, \triangle{TFC}$, $$\frac{\sin \angle{ETC}}{\sin \angle{FTC}} = \frac{\sin \angle{TEC} \cdot \frac{EC}{TC}}{\sin \angle{TFC} \cdot \frac{FC}{TC}} = \frac{\sin \angle{TEC}}{\sin \angle{TFC}} \cdot \frac{EC}{FC}.$$Thus, it suffices to show that $$\frac{\sin \angle{THS}}{\sin \angle{TIS}} \cdot \frac{AF}{BE} = \frac{\sin \angle{TEC}}{\sin \angle{TFC}}.$$Now note that if $A', B'$ are the $A-, B-$ antipodes wrt $(ABC)$, and $P = HS \cap CB', Q = IS \cap CA'$, then $\triangle{PHB'} \sim \triangle{AFA'}, \triangle{QIA'} \sim \triangle{BEB'}$. It suffices to show that $\frac{\sin \angle{PHE}}{\sin \angle{PEH}} \cdot \frac{AF}{BE} = \frac{\sin \angle{QIF}}{\sin \angle{QFI}}$, but since $PH = HB' \cdot \frac{AF}{A'F}$ and $PB' = HB' \cdot \frac{AA'}{A'F}$, $\frac{\sin \angle{PHE}}{\sin \angle{PEH}} = \frac{PE}{PH} = \frac{HB' \cdot \frac{AA'}{A'F} + B'E}{HB' \cdot \frac{AF}{A'F}} = \frac{HB' \cdot AA' + B'E \cdot A'F}{HB' \cdot AF}$. $HB' = IA'$, so the numerator and the $HB'$ in the denominator are the same for $\frac{\sin \angle{QIF}}{\sin \angle{QFI}}$, so it clearly holds. $\square$