2. In an acute-angled triangle $A B C$ with an angle of $45^{\circ}$ at vertex $A$, altitudes $A D, B E$, and $C F$ are drawn. Ray $E F$ intersects line $B C$ at point $X$. It turns out that $A X \| D E$. Find the angles of triangle $A B C$.
Answer: 45, 60, 75. From the condition, it follows that $A F=F C$. From the inscribed angles $\angle C D E=45$, and now from the parallelism $\angle A X D=45$. In triangle $A X C$, point $F$ turned out to be the center of the circumscribed circle (equidistant from vertices $A$ and $C$ and the central angle is twice the inscribed angle). Therefore, $2 \angle A C B=$ $\angle A F X=\angle B F E$, which together with the inscribed quadrilateral $B F E C$ gives $\angle A C B=60^{\circ}$.