The lengths of the sides of $ABC$ denote by $a, b, c$. Let $I$ be the center of the incircle, $r$ its radius, $M$ the midpoint of $BC$. Denote by $r_a$ the radius of the excircle touching $BC$. Let $L$ be the point of tangency of excircle with $BC$. $\omega_0$ be the circle with center $O$, passing through $B$ and $C$, touching $\omega$. W.l.o.g. assume that $b \ge c$.

Firstly, $SK$ is a bisector of the angle $CSB$, since the homothety with center $S$, which transforms the incircle to $\omega_0$, transforms $BC$ to some line touching $\omega_0$ at $E$ the midpoint of $BC$.

It is well-known fact that $K$ and $L$ are symmetric with respect to $M$. So, if we prove that $ME = I_aL/2$ then we can immediately conclude that $K, E, I_a$ are collinear.

Let $OT \perp IK$, $OM = |x|$, where $x \ge 0$, if $O$ lies below $BC$, and $x < 0$ otherwise. Let $R$ be the radius of $\omega_0$. Then $ME = R - x$. Consider the right triangle $OTI$, $OI = R - r$, $OT = (b - c)/2$, $IT = r - x$. By Pythagorean theorem

$$(1) \qquad (R-r)^2 = (r-x)^2 + (b-c)^2/4$$
From the right triangle $OMB$ we have $R^2 - x^2 = a^2/4$. Then using (1) we get $ME = R-x = (p-b)(p-c)/(2r)$, where $p$ is a semiperimeter of $ABC$. Then it is sufficient to prove that
$$(2) \qquad r_a = \frac{(p-b)(p-c)}{r}$$
We have $r = S/p$. On the other hand, $S = r_a(p-a)$. Then (2) true iff $S^2 = p(p-a)(p-b)(p-c)$, which is Heron's formula.