1. Since $BC \parallel EF$, and circles $K$, $L$ are tangent to $EF$ at points $E$, $F$ respectively, we know that points $E$, $F$ are respectively the midpoints of arc $CDE$, arc $BFD$. We obtain $BF = DF$, $CE = DE$. Therefore,
$$\angle EFD = \angle FDB = \angle DBF = \angle EFI.$$
Similarly, $\angle FED = \angle FEI$. From this we know $\triangle DEF \sim \triangle IEF$.

2. $A$, $E$, $I$, $F$ are concyclic:
$$\begin{align*} \angle EAF &= \angle (AE, AD) + \angle (AD, AF) = \angle (CE, CD) + \angle (BD, BF) \\ &= \angle (CI, BC) + \angle (BC, BI) = \angle (CI, BI) \\ &= \angle (EI, FI) = \angle EIF. \end{align*}$$

$$\begin{align*} \angle AEQ &= \angle FEQ - \angle FEA = \angle EDQ - \angle FIA \\ &= \angle EDF - \angle FIA = \angle FIE - \angle FIA = \angle AIE. \end{align*}$$

3. Let $AD$ intersect $EF$ at $N$. Then $NF^2 = NA \cdot ND = NE^2$. So $NF = NE$, and $AD$ is the median of $\triangle DEF$. Since $\triangle DEF \cong \triangle IEF$, and $\angle ADF = \angle AEQ = \angle AIE$, $\angle ADE = \angle AEF = \angle AIF$, $AI$ is the $A$-symmedian of $\triangle EIF$.
Since $EQ$, $FR$ are the tangent lines of $K$ at $E$, $F$, the $A$-symmedian of $\triangle EIF$ passes through the intersection point $M$ of $EQ$, $FR$, that is, $M$ lies on line $AI$. Q.E.D.