$EN \cap FM = X$ and $EC \cap FB = Y$. Prove that $XY$ passes through a fixed point. 
$E$ and $F$ are points respectively on segments $AB$ and $AC$ such that $EF \parallel BC$. The tangents of $(AEF)$ at points $E$ and $F$ cuts $BC$ resepctively at $M$ and $N$. $EN \cap FM = X$ and $EC \cap FB = Y$. Prove that $XY$ passes through a fixed point.


I have a prediction in which $XY$ intersects $BC$ at fixed point $D$. Perhaps where $$\dfrac{DM}{DB} \left(= \dfrac{DX}{DY}\right) = \dfrac{DN}{DC}$$. But I am uncertain.
 A: Let $K$ be the circumcenter of $\triangle AEF$, and define $a:=|BC|$, $b:=|CA|$, $c:=|AB|$.

Since $\angle AKF = 2\angle B$ and $\angle AKE=2\angle C$ (by the Inscribed Angle Theorem), and since $\overline{EM}\perp\overline{EK}$ and $\overline{FN}\perp\overline{FK}$, a little angle chasing gives that $\angle BEM = \angle C$ and $\angle CFN = \angle B$. Consequently, 
$$\triangle MBE\sim\triangle ABC\sim\triangle NFC \tag{1}$$
Since $\overline{EF}\parallel\overline{BC}$, we can write $|BE|=\lambda c$ and $|CF|=\lambda b$ for some $\lambda$ (which we should take to be non-zero). Proportions following from $(1)$ give us
$$|BM|=\lambda\frac{c^2}{a} \qquad |CN|=\lambda\frac{b^2}{a} \tag{2}$$
Borrowing an insight from @liaombro's (currently-deleted) answer, the parallelism (and consequent proportions from $\triangle YBC\sim\triangle YFE$) implies that $Y$ is the center of a homothety carrying $\overline{BC}$ to $\overline{FE}$. Likewise, $X$ is the center of a homothety carrying $\overline{FE}$ to $\overline{MN}$. The product of these transformations is a homothety that carries $\overline{BC}$ to $\overline{MN}$. Since $B$, $C$, $M$, $N$ are collinear, the center (say, $Z$) of that homothety must lie on $\overleftrightarrow{BC}$; by the result linked by @liaombro, $Z$ also lies on $\overleftrightarrow{XY}$. We show that $Z$ is the target fixed point as follows:

$$\underbrace{\frac{|BZ|}{|MZ|} = \frac{|CZ|}{|NZ|}}_{\text{homothety through } Z} \quad\to\quad\frac{|BZ|}{|BZ|-\lambda b^2/a}=\frac{|CZ|}{|CZ|-\lambda c^2/a}\quad\to\quad\frac{|BZ|}{|CZ|}=\frac{b^2}{c^2} \tag{$\star$}$$

Since the final ratio is independent of $\lambda$, the result is shown. $\square$
