Let triangle $ABC$ with $AB < AC$ have incenter $I$ and be circumscribed in $\odot O$. Let $M$ and $N$ be the midpoints of arc $\widehat{BAC}$ and $\widehat{BC}$, respectively. Point $D$ lies on $\odot O$ such that $AD \parallel BC$, and $E$ is the tangency point of the $A$-excircle of $\triangle ABC$. Point $F$ is in $\triangle ABC$ such that $FI \parallel BC$ and $\angle BAF = \angle EAC$. Extend $NF$ to meet $\odot O$ at $G$, and extend $AG$ to meet line $IF$ at $L$. Let line $AF$ and $DI$ meet at $K$. We aim to prove that $ML \perp NK$.

To prove this, consider the following steps:

1. Claim: $G, I, P$ are collinear, where $P$ is the intersection of $AE$ with $\odot O$.
- Proof: Redefine $G'$ as the intersection of $\odot O$ with line $PI$ (other than $P$). Let $F'$ be the intersection of $NG'$ with $AT$. By applying Pascal's theorem on hexagon $ATPG'NN$, we get $IF' \parallel BC$. This implies $F = F'$ and $G = G'$.

2. Claim: $H, F, P$ are collinear, where $H$ is the intersection of $NK$ with $\odot O$.
- Proof: Let $F'$ be the intersection of $HP$ with $AT$. We need to show $F'I \parallel AD$. Using the cross-ratio and the Angle Bisector Theorem, we get:
$$\frac{F'K}{F'A} = \frac{TK}{PA} = \frac{TK}{TD} = \frac{KI}{ID},$$
which implies $F'I \parallel AD \parallel BC$.

3. Claim: $L', F, I$ are collinear, where $L'$ is the intersection of $MH$ with $AG$.
- Proof: Using Pascal's theorem on hexagon $TAGPHM$, we get $F, L', I$ collinear.

Since $L' = L$, we have shown that $ML \perp NK$.

Thus, the proof is complete. $\boxed{\text{ML} \perp \text{NK}}$.