Assume for the sake of contrary such that $\omega$ doesn't contain $I$ in it's interior. Let $P$ is the center of $\omega$. The lines passing $P$ perpendicular to $AB,BC,CA$ are on the segments $AB,BC,CA$. Thus we conclude that $P$ lies inside the circumcircle of $\triangle{ABC}$. Easy to see that $PI>d(P,AB),d(P,BC),d(P,CA)$, and the region of $P$ which satisfies $d(P,AB)>PI$ is in the shape of a parabola. Similarly, so is $d(P,BC),d(P,CA)$. Let me call these parabolas $P_a,P_b,P_c$. Obviously $P_b\cap P_c$ intersects on $AI$. Let $AI\cap(ABC)$ be $D$. Wlog assume $P$ exists in the opposite direction from $I$ with $A$, and it follows that $D$ is outside $P_b,P_c$. But it is famous that $DB=DI=DC$, contradiction.

there is a lemma which states that if we take two isognal congruent points and draw perpendicular lines from them to the sides of the triangle we will have a cyclic hexagon. this is basically this question

Let $O$ be the center of $\omega$. Keep dilating $\omega$ through $O$ by scale $<1$ until $\omega$ touches one side of $\triangle ABC$, say it touches side $AB$ at $T$. Note that it suffices to show $I$ (incenter of $\triangle ABC$) lies inside this new $\omega$. Now Dilate $\omega$ through $T$ by scale $<1$ until $\omega$ touches one of $\overline{BC},\overline{CA}$. Say $\omega$ touches side $AC$ at $U$. Note that it suffices to show $I$ lies inside this new circle $\omega$. Also, note that points $T,U$ must lie on segments $AB,AC$, respectively and $\omega$,$\overline{BC}$ have a common point. Now, for any real $k$, let $\mathcal A_k$ denote the homthety at $A$ with scale $k$. It suffices to show there exist $\alpha \le 1$ and $\beta > 1$ such that $I$ lies inside $\mathcal A_{\alpha}(\omega)$ ; segment $AI$ and $\mathcal A_{\beta} (\omega)$ have a common point. Note that for some such $\alpha$, $\omega$ is send to incircle (as $\omega$ cuts segment $BC$ have a common point). So this is not a issue for us. Now, for some $\beta$, either $T$ goes to $B$ or $U$ goes to $C$ (as $T,U$ must lie on segments $AB,AC$, respectively). In particular, it suffices to prove the following lemma: Lemma: Given $\triangle ABC$ with incenter $I$ and $AB \ge AC$, let $\omega$ be the circle passing through $B$ and tangent to both of $\overline{AB},\overline{AC}$. Prove that segment $AI$ and $\omega$ have a common point. proof: Let $\omega$ touch segment $AC$ at $C'$ and $I'$ be the incenter of $\triangle ABC'$. It is not hard to see that $\omega$ passes through $I'$, and we also see that $I$ lies on segment $AI'$. We are done! $\square$

Dilate twice such that $\omega$ is tangent to WLOG AB and AC. Then since $AI$ is longer than when it is projected, the result follows easily.

arshiya381 wrote: there is a lemma which states that if we take two isognal congruent points and draw perpendicular lines from them to the sides of the triangle we will have a cyclic hexagon. this is basically this question this is called the nine point circle

TheHazard wrote: Dilate twice such that $\omega$ is tangent to WLOG AB and AC. Then since $AI$ is longer than when it is projected, the result follows easily. you could also just draw the two pedal triangles and win immediately