12.026. A rhombus is inscribed in a parallelogram with sides $a$ and $b(a<b)$ and an acute angle $\alpha$; two of its vertices coincide with the midpoints of the larger sides of the parallelogram, the other two lie on the smaller sides (or their extensions). Find the angles of the rhombus.
Solution.

Let $D C=b, B C=a, \angle A D C=\alpha=\angle A K O ; N E F M$ be a rhombus, $K O=\frac{D C}{2}=\frac{b}{2}$ (Fig. 12.29), $O F=\frac{B C}{2}=\frac{a}{2}$. From $\triangle E O F \quad \angle E O F=90^{\circ}$, $\operatorname{tg} \angle O E F=\frac{O F}{E O} \cdot$ From $\triangle K E O \quad \angle K E O=90^{\circ}(A D \| N F, E O$ - transversal $\left.\angle F O E=\angle K E O=90^{\circ}\right), \quad E O=\frac{b}{2} \cdot \sin \alpha$, therefore, $\quad \operatorname{tg} \angle O E F=\frac{a \cdot 2}{2 b \sin \alpha} ;$ $\angle N E F=2 \angle O E F=2 \operatorname{arctg} \frac{a}{b \sin \alpha}, \angle E F M=\pi-2 \operatorname{arctg} \frac{a}{b \sin \alpha}$.

Answer: $2 \operatorname{arctg} \frac{a}{b \sin \alpha} ; \pi-2 \operatorname{arctg} \frac{a}{b \sin \alpha}$.