It’s not too hard. Let $M, N, P$ be the midpoints of $AC, AB, BC$, respectively, and let $BI$ and $CI$ intersect the circle with diameter $BC$ at $L$ and $K$, respectively. It is well known that $\{K \}= PN \cap EF$ and $\{L\}=PM \cap EF $. Take $Z$ as the intersection of $EF$ and $MN$. Then we have to prove that $Z$ lies on the radical axis of the circle with diameter $BC$ and $\odot(I)$. That is equivalent with $ZE \cdot ZF=ZK \cdot ZL$. But this is easy to prove because $\triangle{ZML} \sim \triangle{ZNF}$ and $\triangle{ZME} \sim \triangle{ZNK}.$

Let $\omega$ touch $BC$ at $D$ and $I$ be its center ; line through $A$ parallel to $BC$ meet $EF$ at $S$ ; $T = EF \cap A\text{-midline}$ ; $U = EF \cap BC $ ; $V,M$ be midpoint of segments $DU,BC$, respectively. We want to show $T$ lies on $\ell$, the radical axes of $\omega, \odot(BC)$. Note that $V$ lies on $\ell$ since $$ VD^2 = VU^2 = VB \cdot VC $$where we use $(B,C ; D,U) = -1$. Since $\ell \perp IM$ and $V \in \ell$, so it suffices to show $VT \perp IM$. It is well known that polar of $M$ wrt $\omega$ passes through $S$. It follows $SD$ is the polar of $M$ wrt $\omega$, hence $IM \perp SD$. Note $T$ is the midpoint of segment $US$. So line $VT$ is just the image of line $DS$ under homothety through $U$ with scale $1/2$. Thus $VT \parallel DS$. It follows $VT \perp IM$, completing the proof. $\blacksquare$

A very nice problem. Let the incircle touch $BC$ at $D$, and let $M, N, P$ be the midpoints of $BC, CA, AB$. Let $B_1 = MN \cap EF$, $B_2 = DE \cap NP$, and define $C_1, C_2$ similarly. By Iran Incenter, $B_1, C_1 \in (BC)$, $B_2 \in BB_1$, and $C_2 \in CC_1$. Hence by Reim $B_1C_1C_2B_2$ and $B_2EFC_2$ are both cyclic, so $EF \cap NP$ has equal power wrt $\omega$ and $(BC)$, as desired.