1. [5 points] The altitudes $C F$ and $A E$ of an acute triangle $A B C$ intersect at point $H$. Points $M$ and $N$ are the midpoints of segments $A H$ and $C H$ respectively. It is known that $F M=2, E N=5$, and $F M \| E N$. Find $\angle A B C$, the area of triangle $A B C$, and the radius of the circumscribed circle around it. Answer: $\angle A B C=60^{\circ}, S_{\triangle A B C}=36 \sqrt{3}, R=2 \sqrt{13}$.
Solution. The median of a right triangle drawn from the vertex of the right angle is equal to half the hypotenuse. Since $F M$ and $E N$ are medians of triangles $A F H$ and $C E H$, then $H M=A M=F M=2$, $C N=E N=H N=5$. Let $\angle A H F=\angle C H E=\alpha$. Then $\angle E H F=180^{\circ}-\alpha, \angle E B F=\alpha$ (since in quadrilateral $B F H E$ angles $E$ and $F$ are right). Next, we find: $\angle M F H=\alpha$ (triangle $M F H$ is isosceles), $\angle N E H=\alpha$ (triangle $N E H$ is isosceles), $\angle E N H=180^{\circ}-2 \alpha$. Since $E N \| F M$, angles $M F H$ and $N E H$ are equal as alternate interior angles, from which $\alpha=180^{\circ}-2 \alpha$ and $\alpha=60^{\circ}$. Thus, $\angle A B C=\alpha=60^{\circ}$.

In triangle $A C H$, it is known that $C H=10, A H=4, \angle A H C=120^{\circ}$. By the cosine theorem, we find that $A C^{2}=10^{2}+4^{2}-2 \cdot 10 \cdot 4 \cdot \cos 120^{\circ}=100+16+40=156, A C=2 \sqrt{39}$. Therefore, the radius of the circumscribed circle around triangle $A B C$ is $\frac{A C}{2 \sin \angle A B C}=\frac{2 \sqrt{39}}{\sqrt{3}}=2 \sqrt{13}$.

Triangle $B C F$ is a right triangle with an angle of $60^{\circ}$ at vertex $B$, and $H F=2, C F=C H+$ $H F=12$. Thus, $B C=\frac{C F}{\sin 60^{\circ}}=\frac{12}{0.5 \sqrt{3}}=8 \sqrt{3}$. In addition, $A H=4, H E=5, A E=9$, so $S_{\triangle A B C}=\frac{1}{2} B C \cdot A E=\frac{1}{2} \cdot 8 \sqrt{3} \cdot 9=36 \sqrt{3}$.