Problem 1141

4. As shown in Figure 10, let $H$ be the orthocenter of $\triangle ABC$, $P$ be the intersection of $BE$ and $CF$, and $PH$ intersect the perpendicular bisector of $BC$ at point $Z$.
$$\begin{array}{l} \text{By } \angle HBP = \angle ABH - \angle ABP \\ = 90^{\circ} - \angle BAC - \angle ABP \\ = 90^{\circ} - \angle BEC = \angle JBC \\ \Rightarrow BH \text{ and } BJ \text{ are isogonal lines of } \angle PBC. \end{array}$$

Similarly, $CH$ and $CJ$ are isogonal lines of $\angle PCB$.
Thus, $H$ and $J$ are a pair of isogonal conjugates of $\triangle BPC$.
Therefore, $\angle HBP = \angle JPC$.
Since $ZB = ZC$, $JF = JE$, and $\triangle PFE \sim \triangle PBC$,
then $\triangle PFE \cup \{J\} \sim \triangle PBC \cup \{Z\}$
$\Rightarrow \triangle JEF \sim \triangle ZCB$.
Let $BH$ intersect $AC$ at point $B'$, and $CH$ intersect $AB$ at point $C'$. Then
$$\begin{array}{l} P_{(BE)}^{H} = -HB \cdot HB' = -HC \cdot HC' = P_{(CF)}^{H}, \\ P_{(BE)}^{P} = -PB \cdot PE = -PC \cdot PF = P_{(CF)}^{P}, \end{array}$$

where $P_{(BE)}^{H}$ represents the power of point $H$ with respect to the circle with diameter $BE$.

Thus, point $Z$ lies on line $HP$, which is the radical axis of the circles with diameters $BE$ and $CF$.
Similarly, points $X$ and $Y$ also lie on line $HP$.
Therefore, $XY$ passes through the orthocenter of $\triangle ABC$.