12.109. The height of a regular triangular prism is $H$. A plane passing through the midline of the lower base and the parallel side of the upper base forms an acute dihedral angle $\alpha$ with the plane of the lower base. Find the area of the section formed by this plane.
## Solution.

Let $A B C A_{1} B_{1} C_{1}$ be a regular triangular prism, $A A_{1}=H$, $A N=N B, B M=M C,\left(\left(A_{1} C_{1} M\right) ;(A B C)\right)=\alpha$ (Fig. 12.109). Draw $B E \perp \dot{A} C$ and $E F \| A A_{1} ; \angle F E B=\alpha$ as the linear angle of the dihedral angle. Let $B E \cap M N=O$; then $F O \perp M N$ (by the theorem of three perpendiculars). The section $A_{1} N M C_{1}$ is a trapezoid, since $M N \| A C$. From $\triangle F E O$, we find $F O=\frac{F E}{\sin \alpha}=\frac{H}{\sin \alpha}, E O=H \operatorname{ctg} \alpha$, hence $B E=2 H \operatorname{ctg} \alpha$. From $\triangle B E C$ we get $E C=M N=B E \operatorname{ctg} 60^{\circ}=2 H \operatorname{ctg} \alpha \cdot \frac{\sqrt{3}}{3}=\frac{2 H \sqrt{3} \operatorname{ctg} \alpha}{3}$; thus, $A C=A_{1} C_{1}=\frac{4 H \sqrt{3} \operatorname{ctg} \alpha}{3}$. We have

$$
S_{A_{1} N M C_{1}}=\frac{A C+M N}{2} \cdot F O=H \sqrt{3} \operatorname{ctg} \alpha \cdot \frac{H}{\sin \alpha}=\frac{H^{2} \sqrt{3} \operatorname{ctg} \alpha}{\sin \alpha} .
$$

Answer: $\frac{H^{2} \sqrt{3} \operatorname{ctg} \alpha}{\sin \alpha}$.